JEE Physics Kinematics MCQs Set 01

Answer: (b) 2r, \( \pi r \)

Explanation:
1. Displacement is the straight line between the end points, which is a diameter, 2r.
2. Distance is the length of the semicircular path, \( \pi r \).

Teacher's Note:
1. Displacement is a vector between initial and final positions; distance is a scalar path length.
2. Distance is never less than the magnitude of displacement.

Answer: (d) \( 10\sqrt3 \) m

Explanation:
1. The space diagonal of a cube of side a is \( a\sqrt3 \).
2. So the displacement is \( 10\sqrt3 \) m, about 17.3 m.

Teacher's Note:
1. Displacement depends only on the end points, not on the path flown.
2. Space diagonal = square root of the sum of the squares of the three edges.

Answer: (c) 60 km/h

Explanation:
1. For equal halves, \( \frac{2}{48} = \frac1{40} + \frac1v \).
2. \( \frac1v = \frac1{24} - \frac1{40} = \frac{1}{60} \), so v = 60 km/h.

Teacher's Note:
1. The distance (2000 m) cancels out because the two halves are equal.
2. Use time = distance/speed for each half.

Answer: (c) speed 10 m/s, velocity zero

Explanation:
1. Average speed \( = \frac{2\pi\times100}{62.8} = 10 \) m/s.
2. After a complete lap the displacement is zero, so the average velocity is zero.

Teacher's Note:
1. Speed uses the path length, velocity uses the displacement.
2. Zero average velocity does not mean the car was at rest.

Answer: (d) S₂ = 3S₁

Explanation:
1. \( S_1 = \tfrac12a(10)^2 = 50a \) and the distance in 20 s is \( 200a \).
2. So \( S_2 = 200a - 50a = 150a = 3S_1 \).

Teacher's Note:
1. Equal successive intervals from rest give distances 1 : 3.
2. Total distance in time 2t is four times that in time t.

Answer: (b) 2 s

Explanation:
1. From rest \( s = \tfrac12at^2 \), so \( t \propto \sqrt s \).
2. For s/4 the time is \( \frac{4}{\sqrt4} = 2 \) s.

Teacher's Note:
1. Time is proportional to the square root of the distance from rest.
2. A quarter of the distance takes half the time.

Answer: (c) 5 : 7

Explanation:
1. Distance in the n-th second \( \propto (2n-1) \).
2. For n = 3 and 4 the ratio is 5 : 7.

Teacher's Note:
1. Odd-number rule again.
2. The later second always covers more distance for an accelerating body.

Answer: (a) t₁ = 2.414 t₂

Explanation:
1. Time for first half: \( t_1 = \sqrt{h/g} \). Time for the whole fall: \( \sqrt{2h/g} \).
2. So \( t_2 = (\sqrt2 - 1)\sqrt{h/g} \) and \( \frac{t_1}{t_2} = \frac{1}{\sqrt2-1} = \sqrt2+1 = 2.414 \).

Teacher's Note:
1. The body moves faster in the second half, so it takes less time.
2. Subtract times measured from the start.

Answer: (b) its velocity is zero and its acceleration is the acceleration due to gravity

Explanation:
1. At the top the velocity reverses direction so it is momentarily zero.
2. Gravity still acts, so the acceleration is g downward, the same as at every other point.

Teacher's Note:
1. Zero velocity does not mean zero acceleration.
2. The acceleration of a projectile is g throughout the flight.

Answer: (d) 60 m

Explanation:
1. Total flight time 8 s, so the highest point is reached at 4 s and u = 40 m/s.
2. The two crossings are symmetric about 4 s, at 2 s and 6 s.
3. \( h = 40(2) - \tfrac12(10)(2)^2 = 60 \) m.

Teacher's Note:
1. Use the symmetry of up and down motion about the highest point.
2. Check with t = 6 s: 240 − 180 = 60 m.

Answer: (c) 2

Explanation:
1. While going up, the deceleration is 10 + 6 = 16 m/s². While coming down, the acceleration is 10 − 6 = 4 m/s².
2. For the same height \( \tfrac12(16)t_1^2 = \tfrac12(4)t_2^2 \), so \( t_2/t_1 = 2 \).

Teacher's Note:
1. Drag opposes motion, so it adds on the way up and subtracts on the way down.
2. The descent takes longer than the ascent.

Answer: (c) \( \sqrt{\frac23} \)

Explanation:
1. Upward deceleration = 12 m/s²; downward acceleration = 8 m/s².
2. \( \tfrac12(12)t_1^2 = \tfrac12(8)t_2^2 \Rightarrow \frac{t_1}{t_2} = \sqrt{\frac{8}{12}} = \sqrt{\frac23} \).

Teacher's Note:
1. The height is common to both journeys.
2. Ascent is quicker than descent when drag acts.

Answer: (d) \( v\sqrt{\frac{mg-F}{mg+F}} \)

Explanation:
1. Going up: \( h = \frac{mv^2}{2(mg+F)} \).
2. Coming down: \( v'^2 = \frac{2(mg-F)h}{m} = v^2\frac{mg-F}{mg+F} \).

Teacher's Note:
1. The drag reduces the return speed below v.
2. Use the work-energy theorem separately for the upward and downward journeys.

Answer: (b) 9 : 1

Explanation:
1. Time of ascent is 5 s. Distance in the first second = 50 − 5 = 45 m.
2. The last second of the upward motion is the first second of a fall from the top: 5 m. Ratio = 45 : 5 = 9 : 1.

Teacher's Note:
1. Reverse the last second of motion and treat it as starting from rest.
2. Odd-number rule: 9 : 7 : 5 : 3 : 1.

Answer: (c) 20 m/s

Explanation:
1. Since \( v\,dv = a\,dx \), \( \tfrac12(v^2-u^2) \) equals the area under the a-x graph.
2. Area \( = \tfrac12(30)(10) = 150 \), so \( v^2 = 100 + 300 = 400 \) and v = 20 m/s.

Teacher's Note:
1. The area under an a-x graph gives half the change in v².
2. Do not confuse it with the a-t graph, whose area gives the change in v.

Answer: (a) 2 m/s

Explanation:
1. Between 12 s and 16 s the slope is \( \frac{0-8}{4} = -2 \) m/s.
2. The speed at 12.5 s is the magnitude, 2 m/s.

Teacher's Note:
1. Instantaneous velocity is the slope of the x-t graph.
2. Speed is the magnitude of velocity.

Answer: (b) 1 m/s

Explanation:
1. The speed is maximum when a becomes 0, at x = 1 m.
2. \( \tfrac12v^2 = \text{area} = \tfrac12(1)(1) = 0.5 \), so v = 1 m/s.

Teacher's Note:
1. Area under the a-x graph equals the change in v²/2.
2. The velocity keeps increasing as long as a is positive.

Answer: (d) \( 2t\sqrt{a^2+b^2} \)

Explanation:
1. \( v_x = 2at \) and \( v_y = 2bt \).
2. Speed \( = \sqrt{v_x^2+v_y^2} = 2t\sqrt{a^2+b^2} \).

Teacher's Note:
1. Differentiate each coordinate separately.
2. Combine the components with Pythagoras.

Answer: (c) −9 m/s

Explanation:
1. \( v = 3t^2 - 12t + 3 \) and \( a = 6t - 12 \).
2. a = 0 at t = 2 s, so \( v = 12 - 24 + 3 = -9 \) m/s.

Teacher's Note:
1. Set the second derivative to zero first.
2. Substitute that time into the velocity expression.

Answer: (c) 4 m/s

Explanation:
1. From 0 to 2 s, \( a = 2 - t \): area \( = \tfrac12(2)(2) = 2 \).
2. From 2 to 4 s, a = t − 2: area = 2. Total change in velocity = 4 m/s.

Teacher's Note:
1. The change in velocity is the area under the a-t graph.
2. Split the integral where the modulus changes sign.

Answer: (d) 15.6 m, 8 m/s

Explanation:
1. \( v = \frac{t^4}{4} - t^3 + 5t + C \); v(1) = 6.25 gives C = 2. So v(2) = 4 − 8 + 10 + 2 = 8 m/s.
2. \( s = \frac{t^5}{20} - \frac{t^4}{4} + \frac{5t^2}{2} + 2t + D \); s(1) = 8.30 gives D = 4.
3. s(2) = 1.6 − 4 + 10 + 4 + 4 = 15.6 m.

Teacher's Note:
1. Integrate twice and fix each constant with the given values at t = 1 s.
2. Check the final values by substituting back.

Answer: (b) t = 1 s

Explanation:
1. \( v = 6t^2 - 6t = 6t(t-1) \).
2. v = 0 at t = 0 and t = 1 s.

Teacher's Note:
1. Set dx/dt = 0.
2. A particle at rest momentarily can still have non-zero acceleration.

Answer: (c) b

Explanation:
1. \( \vec a = \frac{d\vec v}{dt} = b\hat j \), independent of time.

Teacher's Note:
1. The acceleration is constant, so the given instant does not matter.
2. Only the j-component of velocity changes.

Answer: (a) The particle has an initial velocity of 5 m/s and a constant acceleration of 12.5 m/s²

Explanation:
1. \( a = v\frac{dv}{dx} = 5\sqrt{1+x}\cdot\frac{5}{2\sqrt{1+x}} = 12.5 \) m/s², a constant.
2. At x = 0, v = 5 m/s, so the initial velocity is 5 m/s.

Teacher's Note:
1. Use a = v dv/dx.
2. Substitute x = 0 for the initial condition.

Answer: (b) 22 m/s

Explanation:
1. For 0 to 5 s, \( v = \int_0^5\sqrt{25-t^2}\,dt \), the area of a quarter circle of radius 5, \( = \frac{25\pi}{4} \).
2. For the next 2 s the increase is \( \frac{3\pi}{8}\times2 = \frac{3\pi}{4} \).
3. Total \( = 7\pi \approx 22 \) m/s.

Teacher's Note:
1. The first integral is the area of a quarter circle.
2. Add the velocity gained in each stage.

Answer: (d) \( 2\hat i + 4\hat j \) m/s

Explanation:
1. \( X = t^2/2 \), so \( v_x = t = 2 \). \( Y = t^4/8 \), so \( v_y = t^3/2 = 4 \).
2. \( \vec v = 2\hat i + 4\hat j \) m/s.

Teacher's Note:
1. Substitute X(t) into the curve to get Y(t).
2. Differentiate each component separately.

Answer: (c) \( \frac{2a}{b} \)

Explanation:
1. \( v\frac{dv}{dx} = a - bx \). Integrating from rest: \( \frac{v^2}{2} = ax - \frac{bx^2}{2} \).
2. The car stops again when v = 0, so \( x = \frac{2a}{b} \).

Teacher's Note:
1. Use a = v dv/dx for acceleration that depends on position.
2. The speed is maximum where a = 0, at x = a/b, which is half of AB.

Answer: (c) 247 m

Explanation:
1. 0 to 5 s: \( \int(2t+13)dt = 25 + 65 = 90 \) m.
2. 5 to 7 s: \( \int(3t+8)dt = 1.5(49-25) + 16 = 52 \) m.
3. 7 to 10 s: \( \int(4t+1)dt = 2(100-49) + 3 = 105 \) m. Total = 247 m.

Teacher's Note:
1. Integrate the speed over each time interval separately.
2. The speed is continuous at t = 5 s and 7 s, which confirms the expressions.