Samacheer Kalvi Class 12 Maths Solutions Chapter 7 Applications of Differential Calculus Exercise 7.1

Question 1. A particle moves along a straight line in such a way that after t seconds its distance from the origin is \( s = 2t^2 + 3t \) metres.
(i) Find the average velocity between \( t = 3 \) and \( t = 6 \) seconds.
(ii) Find the instantaneous velocities at \( t = 3 \) and \( t = 6 \) seconds.
Answer:
The distance from the origin at time \( t \) is given by \( s(t) = 2t^2 + 3t \).
(i) To find the average velocity between \( t = 3 \) and \( t = 6 \) seconds, we use the formula for average velocity:
Average velocity \( = \frac{s(6) - s(3)}{6 - 3} \)
First, calculate \( s(6) \):
\( s(6) = 2(6^2) + 3(6) = 2(36) + 18 = 72 + 18 = 90 \) metres.
Next, calculate \( s(3) \):
\( s(3) = 2(3^2) + 3(3) = 2(9) + 9 = 18 + 9 = 27 \) metres.
Now, substitute these values into the average velocity formula:
Average velocity \( = \frac{90 - 27}{6 - 3} = \frac{63}{3} = 21 \) m/s.
(ii) To find the instantaneous velocity, we need the derivative of the position function, which is the velocity function \( v(t) = s'(t) \).
Given \( s(t) = 2t^2 + 3t \), the derivative is:
\( v(t) = s'(t) = \frac{d}{dt} (2t^2 + 3t) = 4t + 3 \). This function describes how fast the particle is moving at any given time.
Now, we find the instantaneous velocities at \( t = 3 \) seconds and \( t = 6 \) seconds:
At \( t = 3 \) seconds:
\( v(3) = 4(3) + 3 = 12 + 3 = 15 \) m/s.
At \( t = 6 \) seconds:
\( v(6) = 4(6) + 3 = 24 + 3 = 27 \) m/s.
In simple words: First, we found the distance the particle traveled at 3 and 6 seconds. Then, we used these distances to calculate the average speed over that time. For the exact speed at a moment, we used a special math rule called differentiation to find the speed formula, and then put in the times 3 and 6 seconds.

🎯 Exam Tip: Remember that average velocity is the total displacement divided by the total time, while instantaneous velocity is found by taking the derivative of the position function.

Question 2. A camera is accidentally knocked off an edge of a cliff 400 ft high. The camera falls a distance of \( s = 16t^2 \) in t seconds.
(i) How long does the camera fall before it hits the ground?
(ii) What is the average velocity with which the camera falls during the last 2 seconds?
(iii) What is the instantaneous velocity of the camera when it hits the ground?
Answer:
The distance fallen by the camera is given by \( s(t) = 16t^2 \). The cliff height is 400 ft.
(i) The camera hits the ground when the distance fallen \( s(t) \) is equal to the height of the cliff, 400 ft.
So, we set \( s(t) = 400 \):
\( 16t^2 = 400 \)
Divide both sides by 16:
\( t^2 = \frac{400}{16} = 25 \)
Take the square root of both sides (time must be positive):
\( t = \sqrt{25} = 5 \) seconds.
Therefore, the camera falls for 5 seconds before it hits the ground. This calculation helps us determine the total duration of the fall.
(ii) The camera hits the ground at \( t = 5 \) seconds. The last 2 seconds of the fall would be from \( t = 3 \) seconds to \( t = 5 \) seconds.
We need to find the average velocity during this interval:
Average velocity \( = \frac{s(5) - s(3)}{5 - 3} \)
First, find \( s(5) \):
\( s(5) = 16(5^2) = 16(25) = 400 \) ft.
Next, find \( s(3) \):
\( s(3) = 16(3^2) = 16(9) = 144 \) ft.
Now, substitute these values:
Average velocity \( = \frac{400 - 144}{2} = \frac{256}{2} = 128 \) ft/sec.
(iii) To find the instantaneous velocity, we differentiate the position function \( s(t) \) to get the velocity function \( v(t) \).
Given \( s(t) = 16t^2 \), the velocity function is:
\( v(t) = s'(t) = \frac{d}{dt} (16t^2) = 32t \).
The camera hits the ground at \( t = 5 \) seconds. So, we find the instantaneous velocity at \( t = 5 \):
\( v(5) = 32(5) = 160 \) ft/sec.
In simple words: We first figured out how long it took for the camera to drop 400 feet by using the distance formula. Then, to find the average speed in the last two seconds, we calculated the distance it fell between 3 and 5 seconds and divided by 2. Finally, to find the speed at the exact moment it hit the ground, we used a special speed formula and put in the time it took to fall.

🎯 Exam Tip: Pay attention to whether the question asks for average velocity (change in displacement over time) or instantaneous velocity (derivative of displacement at a specific time).

Question 3. A particle moves along a line according to the law \( s(t) = 2t^3 - 9t^2 + 12t - 4 \), where \( t \ge 0 \).
(i) At what times the particle changes direction?
(ii) Find the total distance travelled by the particle in the first 4 seconds.
(iii) Find the particle's acceleration each time the velocity is zero.
Answer:
The position of the particle is given by \( s(t) = 2t^3 - 9t^2 + 12t - 4 \).
(i) The particle changes direction when its velocity is zero and the acceleration is not zero (or the velocity changes sign). First, find the velocity function \( v(t) \) by differentiating \( s(t) \):
\( v(t) = s'(t) = \frac{d}{dt} (2t^3 - 9t^2 + 12t - 4) = 6t^2 - 18t + 12 \).
Set \( v(t) = 0 \) to find the times when the particle might change direction:
\( 6t^2 - 18t + 12 = 0 \)
Divide the entire equation by 6:
\( t^2 - 3t + 2 = 0 \)
Factor the quadratic equation:
\( (t - 1)(t - 2) = 0 \)
This gives two possible times: \( t = 1 \) second or \( t = 2 \) seconds.
To confirm a change in direction, we need to check the sign of velocity before and after these points. For a quadratic \( t^2 - 3t + 2 \), it is positive outside the roots ( \( t < 1 \) or \( t > 2 \)) and negative between the roots ( \( 1 < t < 2 \)). Since the velocity changes sign at both \( t=1 \) and \( t=2 \), the particle changes direction at these times.
(ii) To find the total distance travelled, we need to consider where the particle changes direction. We found this happens at \( t=1 \) and \( t=2 \). We must sum the absolute values of the distances between these points and the end points.
The intervals are \( [0, 1] \), \( [1, 2] \), \( [2, 3] \), and \( [3, 4] \).
First, calculate the position at these times:
\( s(0) = 2(0)^3 - 9(0)^2 + 12(0) - 4 = -4 \)
\( s(1) = 2(1)^3 - 9(1)^2 + 12(1) - 4 = 2 - 9 + 12 - 4 = 1 \)
\( s(2) = 2(2)^3 - 9(2)^2 + 12(2) - 4 = 2(8) - 9(4) + 24 - 4 = 16 - 36 + 24 - 4 = 0 \)
\( s(3) = 2(3)^3 - 9(3)^2 + 12(3) - 4 = 2(27) - 9(9) + 36 - 4 = 54 - 81 + 36 - 4 = 5 \)
\( s(4) = 2(4)^3 - 9(4)^2 + 12(4) - 4 = 2(64) - 9(16) + 48 - 4 = 128 - 144 + 48 - 4 = 28 \)
The total distance travelled is:
\( |s(0) - s(1)| + |s(1) - s(2)| + |s(2) - s(3)| + |s(3) - s(4)| \)
\( = |-4 - 1| + |1 - 0| + |0 - 5| + |5 - 28| \)
\( = |-5| + |1| + |-5| + |-23| \)
\( = 5 + 1 + 5 + 23 = 34 \) metres.
(iii) The particle's acceleration is found by differentiating the velocity function \( v(t) \).
We have \( v(t) = 6t^2 - 18t + 12 \).
The acceleration function \( a(t) = v'(t) = s''(t) \):
\( a(t) = \frac{d}{dt} (6t^2 - 18t + 12) = 12t - 18 \).
We need to find the acceleration at the times when velocity is zero, which are \( t = 1 \) second and \( t = 2 \) seconds.
At \( t = 1 \) second:
\( a(1) = 12(1) - 18 = 12 - 18 = -6 \) m/s\(^2\).
At \( t = 2 \) seconds:
\( a(2) = 12(2) - 18 = 24 - 18 = 6 \) m/s\(^2\).
In simple words: We first found when the particle stopped and turned around by calculating its speed and seeing when it became zero. Then, to get the total distance it moved, we added up all the small distances it covered, even when it went backward. Finally, we found its acceleration (how quickly its speed changed) at the moments it turned around.

🎯 Exam Tip: For total distance, always consider points where velocity is zero, as the particle might change direction. Calculate displacement in each interval and sum their absolute values.

Question 4. If the volume of a cube of side length x is \( v = x^3 \). Find the rate of change of the volume with respect to x when \( x = 5 \) units.
Answer:
The volume of a cube with side length \( x \) is given by \( v = x^3 \).
To find the rate of change of the volume with respect to \( x \), we need to differentiate \( v \) with respect to \( x \):
\( \frac{dv}{dx} = \frac{d}{dx} (x^3) = 3x^2 \).
Now, we need to find this rate of change when \( x = 5 \) units. We substitute \( x = 5 \) into the derivative:
\( \frac{dv}{dx} \Big|_{x=5} = 3(5)^2 = 3(25) = 75 \).
This means that when the side length is 5 units, the volume is increasing 75 times faster than the side length is increasing.
In simple words: The volume of a cube is found by multiplying its side length by itself three times. To see how fast the volume grows as the side length grows, we use a math tool called differentiation. We found a formula for this growth, and when the side is 5 units long, the volume is growing at a rate of 75 units per unit of side length.

🎯 Exam Tip: Remember that "rate of change" almost always means finding the derivative. Pay close attention to "with respect to what" (e.g., \( \frac{dv}{dx} \) vs. \( \frac{dv}{dt} \)).

Question 5. If the mass \( m(x) \) (in kilograms) of a thin rod of length \( x \) (in metres) is given by \( m(x) = \sqrt{3x} \), then what is the rate of change of mass with respect to the length when it is \( x = 3 \) and \( x = 27 \) metres.
Answer:
The mass of the thin rod is given by \( m(x) = \sqrt{3x} \).
We can rewrite this as \( m(x) = \sqrt{3} \cdot \sqrt{x} = \sqrt{3}x^{1/2} \).
To find the rate of change of mass with respect to length, we differentiate \( m(x) \) with respect to \( x \):
\( \frac{dm}{dx} = m'(x) = \frac{d}{dx} (\sqrt{3}x^{1/2}) = \sqrt{3} \cdot \frac{1}{2}x^{(1/2)-1} = \frac{\sqrt{3}}{2}x^{-1/2} = \frac{\sqrt{3}}{2\sqrt{x}} \).
Now, we evaluate this rate of change at the given lengths:
When \( x = 3 \) metres:
\( m'(3) = \frac{\sqrt{3}}{2\sqrt{3}} = \frac{1}{2} \) kg/m.
When \( x = 27 \) metres:
\( m'(27) = \frac{\sqrt{3}}{2\sqrt{27}} = \frac{\sqrt{3}}{2\sqrt{9 \cdot 3}} = \frac{\sqrt{3}}{2 \cdot 3\sqrt{3}} = \frac{\sqrt{3}}{6\sqrt{3}} = \frac{1}{6} \) kg/m.
This rate tells us how much the mass increases for each small increase in length at that specific point.
In simple words: We have a formula for how heavy a rod is based on its length. To find out how fast the weight changes as the length changes, we used a special math rule to get a "rate of change" formula. Then we put in the given lengths to find the exact rate for each.

🎯 Exam Tip: When dealing with square roots, it's often helpful to rewrite them using fractional exponents (e.g., \( \sqrt{x} = x^{1/2} \)) before differentiating.

Question 6. A stone is dropped into a pond causing ripples in the form of concentric circles. The radius r of the outer ripple is increasing at a constant rate of 2 cm per second. When the radius is 5 cm find the rate of changing of the total area of the disturbed water?
Answer:
Let \( r \) be the radius of the circular ripple. The rate at which the radius is increasing is given as \( \frac{dr}{dt} = 2 \) cm/sec.
The area \( A \) of a circle is given by the formula \( A = \pi r^2 \).
We need to find the rate of change of the area with respect to time, \( \frac{dA}{dt} \). We use the chain rule to differentiate \( A \) with respect to \( t \):
\( \frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} \).
First, find \( \frac{dA}{dr} \):
\( \frac{dA}{dr} = \frac{d}{dr}(\pi r^2) = 2\pi r \).
Now substitute this back into the chain rule formula:
\( \frac{dA}{dt} = (2\pi r) \cdot \frac{dr}{dt} \).
We are given that \( \frac{dr}{dt} = 2 \) cm/sec. We need to find \( \frac{dA}{dt} \) when \( r = 5 \) cm.
Substitute the values:
\( \frac{dA}{dt} = 2\pi (5) (2) = 20\pi \) cm\(^2\)/sec.
So, when the radius is 5 cm, the area of the disturbed water is increasing at a rate of \( 20\pi \) cm\(^2\)/sec. This shows how quickly the circular area expands as the ripple grows.
In simple words: We know how fast the ripple's circle gets bigger in radius. We used the formula for the area of a circle and a special math rule called the chain rule to find out how fast the entire area of the ripple is growing when its radius is 5 cm.

🎯 Exam Tip: Problems involving rates of change of geometric figures often require implicit differentiation and the chain rule. Draw a diagram if it helps visualize the relationships between variables.

Question 7. A searchlight is located 5 km from a straight shore line. How fast is the beam moving along the shoreline when it makes an angle of 45° with the shore?
Answer:
Let O be the position of the searchlight, and let A be the point on the shoreline closest to O. So, \( OA = 5 \) km. Let B be the point where the light beam hits the shoreline, and let \( x = AB \) be the distance along the shoreline from A to B. Let \( \theta \) be the angle the beam makes with the line OA.

From the right-angled triangle OAB, we have:
\( \tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{AB}{OA} = \frac{x}{5} \)
So, \( x = 5 \tan \theta \).
We are given that the searchlight completes one revolution in 10 seconds. This means the angle \( \theta \) changes by \( 2\pi \) radians in 10 seconds.
The angular velocity \( \frac{d\theta}{dt} = \frac{2\pi \text{ radians}}{10 \text{ seconds}} = \frac{\pi}{5} \) rad/sec.
We need to find \( \frac{dx}{dt} \) when \( \theta = 45^\circ = \frac{\pi}{4} \) radians.
Differentiate \( x = 5 \tan \theta \) with respect to \( t \) using the chain rule:
\( \frac{dx}{dt} = \frac{d}{d\theta} (5 \tan \theta) \cdot \frac{d\theta}{dt} \)
\( \frac{dx}{dt} = 5 \sec^2 \theta \cdot \frac{d\theta}{dt} \)
Now, substitute \( \theta = 45^\circ \) and \( \frac{d\theta}{dt} = \frac{\pi}{5} \):
Recall that \( \sec 45^\circ = \frac{1}{\cos 45^\circ} = \frac{1}{1/\sqrt{2}} = \sqrt{2} \).
So, \( \sec^2 45^\circ = (\sqrt{2})^2 = 2 \).
\( \frac{dx}{dt} = 5 (2) \left( \frac{\pi}{5} \right) = 10 \left( \frac{\pi}{5} \right) = 2\pi \) km/sec.
Therefore, the beam is moving along the shoreline at a rate of \( 2\pi \) km/sec when the angle is 45 degrees.
In simple words: We used trigonometry to relate the distance the light travels on the shore to the angle of the searchlight beam. Knowing how fast the angle changes, we used a math rule to find out how fast the light spot moves on the shore when the angle is 45 degrees.

🎯 Exam Tip: For problems involving angles and rates, ensure your calculator is in radian mode for trigonometric functions when working with derivatives if the angular velocity is in radians per unit time.

Question 8. A conical water tank with a vertex down of 12 metres height has a radius of 5 metres at the top. If water flows into the tank at a rate of 10 cubic m/min, how fast is the depth of the water increases when the water is 8 metres deep?
Answer:
Let \( H \) be the height of the cone and \( R \) be the radius of the top. Given \( H = 12 \) m and \( R = 5 \) m.
Let \( h \) be the depth of the water at any time \( t \), and \( r \) be the radius of the water surface at depth \( h \).

By similar triangles, we have the ratio \( \frac{r}{h} = \frac{R}{H} \).
Substituting the given values: \( \frac{r}{h} = \frac{5}{12} \).
So, \( r = \frac{5h}{12} \). This equation relates the radius of the water surface to its depth.
The volume \( V \) of water in the cone is given by \( V = \frac{1}{3}\pi r^2 h \).
Substitute the expression for \( r \) into the volume formula:
\( V = \frac{1}{3}\pi \left(\frac{5h}{12}\right)^2 h \)
\( V = \frac{1}{3}\pi \left(\frac{25h^2}{144}\right) h \)
\( V = \frac{25\pi h^3}{432} \).
We are given that water flows into the tank at a rate of \( \frac{dV}{dt} = 10 \) cubic m/min.
We need to find \( \frac{dh}{dt} \) when \( h = 8 \) metres.
Differentiate the volume formula with respect to time \( t \) using the chain rule:
\( \frac{dV}{dt} = \frac{d}{dh} \left(\frac{25\pi h^3}{432}\right) \cdot \frac{dh}{dt} \)
\( \frac{dV}{dt} = \frac{25\pi}{432} (3h^2) \frac{dh}{dt} \)
\( \frac{dV}{dt} = \frac{25\pi h^2}{144} \frac{dh}{dt} \).
Now, substitute the known values: \( \frac{dV}{dt} = 10 \) and \( h = 8 \):
\( 10 = \frac{25\pi (8^2)}{144} \frac{dh}{dt} \)
\( 10 = \frac{25\pi (64)}{144} \frac{dh}{dt} \)
\( 10 = \frac{25\pi \cdot 4}{9} \frac{dh}{dt} \) (since \( \frac{64}{144} = \frac{4 \cdot 16}{9 \cdot 16} = \frac{4}{9} \))
\( 10 = \frac{100\pi}{9} \frac{dh}{dt} \)
Now, solve for \( \frac{dh}{dt} \):
\( \frac{dh}{dt} = \frac{10 \cdot 9}{100\pi} = \frac{90}{100\pi} = \frac{9}{10\pi} \) m/min.
The depth of the water is increasing at the rate of \( \frac{9}{10\pi} \) m/min. This rate shows how quickly the water level rises in the tank.
In simple words: We used similar triangles to relate the water's radius to its depth. Then, we wrote down the formula for the volume of water in the cone. By using a special math rule, we found out how fast the water depth is increasing when water is flowing in at a certain rate and the depth is 8 meters.

🎯 Exam Tip: For related rates problems involving cones or other similar shapes, always establish a relationship between the variables (e.g., radius and height) using similar triangles before differentiating.

Question 9. A ladder 17 metre long is leaning against the wall. The base of the ladder is moving away from the wall at a rate of 5 m/s. When the base of the ladder is 8 metres from the wall, (i) How fast is the top of the ladder moving down the wall? (ii) At what rate, the area of the triangle formed by the ladder, wall, and floor is changing?
Answer:
Let \( x \) be the distance of the base of the ladder from the wall, and \( y \) be the height of the top of the ladder on the wall. The length of the ladder is constant at 17 metres.

By the Pythagorean theorem, \( x^2 + y^2 = 17^2 \).
Given values: \( x = 8 \) m and \( \frac{dx}{dt} = 5 \) m/s.
First, find \( y \) when \( x = 8 \):
\( 8^2 + y^2 = 17^2 \)
\( 64 + y^2 = 289 \)
\( y^2 = 289 - 64 = 225 \)
\( y = \sqrt{225} = 15 \) m.
(i) To find how fast the top of the ladder is moving down, we differentiate the Pythagorean equation with respect to time \( t \):
\( \frac{d}{dt} (x^2 + y^2) = \frac{d}{dt} (17^2) \)
\( 2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0 \)
Divide by 2:
\( x \frac{dx}{dt} + y \frac{dy}{dt} = 0 \).
Substitute the known values: \( x = 8 \), \( y = 15 \), \( \frac{dx}{dt} = 5 \):
\( 8(5) + 15 \frac{dy}{dt} = 0 \)
\( 40 + 15 \frac{dy}{dt} = 0 \)
\( 15 \frac{dy}{dt} = -40 \)
\( \frac{dy}{dt} = -\frac{40}{15} = -\frac{8}{3} \) m/s.
The negative sign indicates that \( y \) is decreasing, meaning the top of the ladder is moving down the wall. The speed at which it moves down is \( \frac{8}{3} \) m/s.
(ii) The area of the triangle formed by the ladder, wall, and floor is \( A = \frac{1}{2}xy \).
To find the rate of change of the area, we differentiate \( A \) with respect to \( t \) using the product rule:
\( \frac{dA}{dt} = \frac{d}{dt} \left(\frac{1}{2}xy\right) = \frac{1}{2} \left( x \frac{dy}{dt} + y \frac{dx}{dt} \right) \).
Substitute the values \( x = 8 \), \( y = 15 \), \( \frac{dx}{dt} = 5 \), and \( \frac{dy}{dt} = -\frac{8}{3} \):
\( \frac{dA}{dt} = \frac{1}{2} \left( 8 \left(-\frac{8}{3}\right) + 15(5) \right) \)
\( \frac{dA}{dt} = \frac{1}{2} \left( -\frac{64}{3} + 75 \right) \)
To add the terms inside the parenthesis, find a common denominator:
\( \frac{dA}{dt} = \frac{1}{2} \left( -\frac{64}{3} + \frac{225}{3} \right) \)
\( \frac{dA}{dt} = \frac{1}{2} \left( \frac{225 - 64}{3} \right) \)
\( \frac{dA}{dt} = \frac{1}{2} \left( \frac{161}{3} \right) \)
\( \frac{dA}{dt} = \frac{161}{6} \approx 26.83 \) m\(^2\)/sec.
The area of the triangle is increasing at approximately \( 26.83 \) m\(^2\)/sec. This means the space enclosed by the ladder, wall, and ground is expanding.
In simple words: We pictured the ladder, wall, and floor as a triangle. Using a math rule about triangles with changing sides, we found how fast the top of the ladder slides down as its base moves away. Then, we calculated how quickly the area of this triangle changes at that exact moment.

🎯 Exam Tip: Always draw a diagram for ladder problems to clearly label variables and establish relationships using the Pythagorean theorem. The sign of the rate of change (positive or negative) indicates whether a quantity is increasing or decreasing.

Question 10. A police jeep, approaching an orthogonal intersection from the northern direction, is chasing a speeding car that has turned and moving straight east. When the jeep is 0.6 km north of the intersection and the car is 0.8 km to the east. The police determine with a radar that the distance between them and the car is increasing at 20 km/hr. If the jeep is moving at 60 km/hr at the instant of measurement, what is the speed of the car?
Answer:
Let the intersection be the origin (0,0).
Let \( y \) be the distance of the jeep from the intersection (north direction).
Let \( x \) be the distance of the car from the intersection (east direction).
Let \( S \) be the distance between the jeep and the car.

From the Pythagorean theorem, \( S^2 = x^2 + y^2 \).
Given values at the instant of measurement:
Car's distance from intersection \( x = 0.8 \) km.
Jeep's distance from intersection \( y = 0.6 \) km.
Rate of increase of distance between them \( \frac{dS}{dt} = 20 \) km/hr.
Speed of the jeep (moving towards the intersection) \( \frac{dy}{dt} = -60 \) km/hr (negative because \( y \) is decreasing).
First, find \( S \) at this instant:
\( S^2 = (0.8)^2 + (0.6)^2 \)
\( S^2 = 0.64 + 0.36 \)
\( S^2 = 1 \)
\( \implies S = 1 \) km (distance is positive).
Now, differentiate the equation \( S^2 = x^2 + y^2 \) with respect to time \( t \):
\( 2S \frac{dS}{dt} = 2x \frac{dx}{dt} + 2y \frac{dy}{dt} \)
Divide by 2:
\( S \frac{dS}{dt} = x \frac{dx}{dt} + y \frac{dy}{dt} \).
Substitute all the known values:
\( 1(20) = (0.8) \frac{dx}{dt} + (0.6)(-60) \)
\( 20 = 0.8 \frac{dx}{dt} - 36 \)
Add 36 to both sides:
\( 20 + 36 = 0.8 \frac{dx}{dt} \)
\( 56 = 0.8 \frac{dx}{dt} \)
Solve for \( \frac{dx}{dt} \):
\( \frac{dx}{dt} = \frac{56}{0.8} = \frac{560}{8} = 70 \) km/hr.
Therefore, the speed of the car at this instant is 70 km/hr. This calculation helps the police understand how fast the car is moving to the east.
In simple words: We used the triangle formed by the jeep, car, and intersection, along with the Pythagorean theorem, to relate their distances. By applying a special math rule for changing speeds, and plugging in all the known speeds and distances, we could figure out how fast the car was going.

🎯 Exam Tip: For problems involving distances and rates in a coordinate plane, always start by drawing a diagram. Remember to assign negative rates for quantities that are decreasing, like the jeep's distance from the intersection.