JEE Physics Motion In A Straight Line MCQs Set 09

Answer: (c)

Explanation:
1. The slope of the x-t graph is the velocity.
2. From 0 to 5 s the slope is positive and decreases to zero at the top of the arch.
3. From 5 s to 10 s the slope is negative and becomes more negative.
4. So velocity falls steadily from a positive value, crosses zero at t = 5 s and goes negative, which is graph (c).

Teacher's Note:
1. Velocity is the slope of the displacement-time graph.
2. A maximum of the x-t graph always means zero velocity at that instant.

Answer: (b) E

Explanation:
1. Velocity is the slope of the displacement-time curve.
2. At C the curve is rising, so velocity is positive.
3. At D the curve is at its maximum, so the slope is zero.
4. At E the curve is falling, so the slope is negative. At F the curve is rising again.

Teacher's Note:
1. Rising curve means positive velocity, falling curve means negative velocity.
2. A turning point (peak or valley) has zero velocity.

Answer: (d) 20 m/s

Explanation:
1. The time to reach the top is half of the total time, so it is 2 s.
2. At the top v = 0, so 0 = u − g × 2.
3. u = 10 × 2 = 20 m/s.

Teacher's Note:
1. Time of ascent equals time of descent for a ball that returns to the same level.
2. Use v = u − gt at the top.

Answer: (b) \( \frac{1}{2}gt^2 \)

Explanation:
1. Reverse the motion: the last t seconds of the ascent look like a fall from rest, starting at the top.
2. Distance in a free fall from rest in time t is \( \frac{1}{2}gt^2 \).
3. So the distance covered is \( \frac{1}{2}gt^2 \), independent of u.

Teacher's Note:
1. Time-reverse the last part of the upward motion to make it a fall from rest.
2. The result does not depend on the initial speed.

Answer: (d) Both Assertion and Reason are false.

Explanation:
1. Average speed is total distance divided by time, while average velocity is displacement divided by time.
2. Distance is always greater than or equal to displacement, so average speed is greater than or equal to the magnitude of average velocity, and they are equal only for straight-line motion without reversal.
3. The Assertion is false, the Reason is true.

Teacher's Note:
1. Distance is a path length, displacement is a straight-line change in position.
2. Because of this, the two averages are not generally equal.

Answer: (d) increase with time

Explanation:
1. Velocity \( v = \frac{dx}{dt} = -a\alpha e^{-\alpha t} + b\beta e^{\beta t} \).
2. Acceleration \( \frac{dv}{dt} = a\alpha^2 e^{-\alpha t} + b\beta^2 e^{\beta t} \), which is always positive.
3. So the velocity increases with time.

Teacher's Note:
1. Differentiate twice and check the sign of the second derivative.
2. A positive acceleration at all times means velocity always increases.

Answer: (a) velocity increases uniformly, stays constant, then increases uniformly again at the same rate

Explanation:
1. The slope of the v-t graph is the acceleration.
2. Positive constant acceleration gives a straight rising line, zero acceleration gives a horizontal line, and the same positive acceleration again gives a rising line with the same slope.
3. This matches option (a), with the particle starting from rest.

Teacher's Note:
1. The slope of a velocity-time graph equals acceleration.
2. Match each phase of the acceleration graph to a line segment.

Answer: (d) \( \left(\frac{3v_0^2}{2}\right)^{1/3} \)

Explanation:
1. Write \( a = v\frac{dv}{dx} = -x^2 \), so \( v\,dv = -x^2\,dx \).
2. Integrate from the origin (speed \( v_0 \)) to the stopping point (speed 0): \( 0 - \frac{v_0^2}{2} = -\frac{x^3}{3} \).
3. So \( x^3 = \frac{3v_0^2}{2} \) and \( x = \left(\frac{3v_0^2}{2}\right)^{1/3} \).

Teacher's Note:
1. When acceleration depends on position, use \( a = v\,dv/dx \).
2. Integrate between the starting speed and zero.

Answer: (c) average speed is u

Explanation:
1. The particle is back at the start, so displacement is zero and average velocity is zero.
2. Maximum height \( H = \frac{u^2}{2g} \) and total time \( T = \frac{2u}{g} \).
3. Average speed = distance/time = \( \frac{2H}{T} = \frac{u}{2} \), so the statement that it is u is incorrect.

Teacher's Note:
1. Average speed uses total path length, average velocity uses displacement.
2. Here the average speed is half of the initial speed.

Answer: (b) a body has a constant speed but it can have a varying velocity

Explanation:
1. Constant velocity means constant magnitude and direction, so speed is also constant, which rules out (a).
2. In uniform circular motion the speed is constant but the direction changes, so velocity varies and there is a centripetal acceleration, which rules out (c).
3. A force perpendicular to motion does no work (circular motion), which rules out (d).
4. Only (b) is correct.

Teacher's Note:
1. Velocity is a vector and speed is its magnitude.
2. Uniform circular motion is the standard example of constant speed with changing velocity.

Answer: (c) \( -2av^3 \)

Explanation:
1. Differentiate with respect to x: \( \frac{dt}{dx} = 2ax + b = \frac{1}{v} \), so \( v = \frac{1}{2ax+b} \).
2. Acceleration \( a_c = v\frac{dv}{dx} \) with \( \frac{dv}{dx} = -\frac{2a}{(2ax+b)^2} = -2av^2 \).
3. So acceleration \( = v \times (-2av^2) = -2av^3 \).

Teacher's Note:
1. Express v in terms of x first, then use \( a = v\,dv/dx \).
2. The result is negative, meaning retardation.

Answer: (c) constant

Explanation:
1. Since \( a = v\frac{dv}{dx} \) and a = kv, we get \( \frac{dv}{dx} = k \).
2. The slope of the v-x graph is therefore the constant k.

Teacher's Note:
1. The slope of a v-x graph is dv/dx.
2. Use \( a = v\,dv/dx \) to connect it with acceleration.

Answer: (c) \( \frac{3v_1v_2}{2v_1+v_2} \)

Explanation:
1. Total distance = 3L.
2. Total time = \( \frac{L}{v_1} + \frac{2L}{v_2} = \frac{Lv_2 + 2Lv_1}{v_1v_2} \).
3. Average speed = \( \frac{3L v_1v_2}{L(2v_1+v_2)} = \frac{3v_1v_2}{2v_1+v_2} \).

Teacher's Note:
1. Average speed is total distance over total time, never the average of speeds.
2. Compute the time for each part separately.

Answer: (d) 1 : 16

Explanation:
1. Using \( v^2 = u^2 - 2as \) with final speed 0, the stopping distance is \( s = \frac{u^2}{2a} \).
2. For the same a, s is proportional to the square of the initial speed.
3. Ratio = \( u^2 : (4u)^2 = 1 : 16 \).

Teacher's Note:
1. Stopping distance is proportional to the square of the speed.
2. Four times the speed means sixteen times the distance.

Answer: (a) 6 m/s²

Explanation:
1. Velocity \( v = 3t^2 - 12t + 9 \).
2. Acceleration \( a = 6t - 12 \).
3. At t = 1 s, a = −6 m/s², so the deceleration is 6 m/s².

Teacher's Note:
1. Differentiate position twice to get acceleration.
2. A negative acceleration while the velocity is positive is a deceleration.

Answer: (b) 0.02 m/s²

Explanation:
1. Time taken = distance / average velocity = 3.06 / 0.34 = 9 s.
2. Acceleration = change in velocity / time = 0.18 / 9 = 0.02 m/s².

Teacher's Note:
1. Average velocity and distance give the time.
2. Then use a = Δv/t.

Answer: (a) uniform acceleration

Explanation:
1. If \( s = kt^2 \), then \( v = \frac{ds}{dt} = 2kt \) and \( a = \frac{dv}{dt} = 2k \).
2. The acceleration is a constant, so the motion has uniform acceleration.

Teacher's Note:
1. Differentiate twice to read off acceleration.
2. A constant result means uniform acceleration.

Answer: (c) a straight line sloping downward with slope −g that crosses the time axis at the top of the flight

Explanation:
1. The velocity changes uniformly because \( v = u - gt \).
2. This is a straight line of slope −g starting at +u.
3. It crosses zero at the top and becomes negative on the way down.

Teacher's Note:
1. Free fall has constant acceleration, so velocity is linear in time.
2. Upward is positive, so the slope is negative.

Answer: (d) 20 m/s

Explanation:
1. The scooterist must cover the 1000 m gap in 100 s, so the relative speed is 1000 / 100 = 10 m/s.
2. The scooterist's velocity = 10 + 10 = 20 m/s.

Teacher's Note:
1. Overtaking uses relative velocity.
2. Gap divided by time gives the required relative speed.

Answer: (d) both 1 s and 3 s

Explanation:
1. Velocity \( v = 3t^2 - 12t + 9 = 3(t-1)(t-3) \).
2. v = 0 at t = 1 s and t = 3 s.

Teacher's Note:
1. Momentary rest means velocity zero.
2. Solve the quadratic obtained from dx/dt.

Answer: (a) 32.7 km/h

Explanation:
1. Let the total distance be 3d.
2. Time = \( \frac{d}{20} + \frac{d}{40} + \frac{d}{60} = \frac{11d}{120} \).
3. Average speed = \( \frac{3d}{11d/120} = \frac{360}{11} \approx 32.7 \) km/h.

Teacher's Note:
1. Equal distances at different speeds need the harmonic mean, not the arithmetic mean.
2. Always compute total time.

Answer: (a) 3 m/s²

Explanation:
1. \( a = v\frac{dv}{dx} = \frac{1}{2}\frac{d(v^2)}{dx} = \frac{1}{2}\frac{d(9-x^2)}{dx} = -x \).
2. The particle can only be where \( 9 - x^2 \ge 0 \), that is |x| ≤ 3.
3. The maximum magnitude of acceleration is 3 m/s² at |x| = 3.

Teacher's Note:
1. Use \( a = \frac{1}{2}\frac{d(v^2)}{dx} \) when v is given as a function of x.
2. The allowed range of x limits the maximum.

Answer: (b) \( \frac{b-a}{t^2} \)

Explanation:
1. For successive equal intervals the difference of distances is \( \Delta s = a_c t^2 \).
2. So \( b - a = a_c t^2 \).
3. Acceleration \( = \frac{b-a}{t^2} \).

Teacher's Note:
1. This is a standard result for constant acceleration.
2. The distance difference over equal intervals equals acceleration times t squared.

Answer: (c) If the speed of a particle is constant, its acceleration must be zero

Explanation:
1. At the top of a vertical throw the velocity is zero but the acceleration is g, so (a) is correct.
2. A body moving at constant velocity has zero acceleration but non-zero velocity, so (b) is correct.
3. Uniform circular motion has constant speed but non-zero acceleration, so (c) is wrong.

Teacher's Note:
1. Zero velocity at an instant does not imply zero acceleration.
2. Constant speed does not imply constant velocity.

Answer: (c) \( \frac{v_1+v_2}{2} \)

Explanation:
1. In time t each, the total displacement = \( v_1t + v_2t \).
2. Total time = 2t.
3. Average velocity = \( \frac{v_1+v_2}{2} \).

Teacher's Note:
1. For equal times the average is the arithmetic mean.
2. For equal distances it would be the harmonic mean.

Answer: (d) \( 10\sqrt{2} \) s

Explanation:
1. The car must gain 200 m on the bus.
2. Relative acceleration = 4 − 2 = 2 m/s², and both start from rest, so \( 200 = \frac{1}{2} \times 2 \times t^2 \).
3. \( t^2 = 200 \), so \( t = 10\sqrt{2} \) s.

Teacher's Note:
1. Work in the frame of the bus and use the relative acceleration.
2. The relative initial velocity is zero.