JEE Physics Motion In A Straight Line MCQs Set 18

Answer: (b) 10 m/s and 20 m/s respectively

Explanation:
1. Total time = 18 + 6 = 24 s. Displacement = OQ = 240 m.
2. Average velocity = 240/24 = 10 m/s.
3. Distance travelled = OP + PQ = 360 + 120 = 480 m, so average speed = 480/24 = 20 m/s.

Teacher's Note:
1. Average velocity uses displacement (a vector), average speed uses total path length.
2. Here the car reverses direction, so the two are different.

Answer: (d) 70 m/s

Explanation:
1. Let the acceleration be a. In the first stage: \( 50^2-10^2=2as\Rightarrow as=1200 \).
2. After reversal the acceleration is −a and the particle must undergo a displacement of −s to return: \( v^2=50^2+2(-a)(-s)=2500+2400=4900 \).
3. \( |v|=70\ \text{m/s} \)

Teacher's Note:
1. Use v² = u² + 2as in each stage and note that the final displacement is the negative of the first stage displacement.
2. The particle is moving back towards the origin when it returns.

Answer: (a) \( 2\alpha v^3 \)

Explanation:
1. \( \frac{dt}{dx}=2\alpha x+\beta=\frac1v \), so \( v=\frac1{2\alpha x+\beta} \).
2. \( a=v\frac{dv}{dx}=v\cdot\left(-2\alpha v^2\right)=-2\alpha v^3 \).
3. Retardation = 2αv³.

Teacher's Note:
1. Write acceleration as v dv/dx when v is given in terms of x.
2. Differentiating t with respect to x gives 1/v directly.

Answer: (a) Bomb from B1 reaches the ground first

Explanation:
1. Each bomb starts with the upward velocity of its balloon, then moves under gravity from the same height h.
2. \( -h=ut-\frac12gt^2 \) with u = v, 2v or 3v (upward).
3. A larger upward initial velocity gives a longer time of flight, so the bomb from the slowest balloon B1 lands first.

Teacher's Note:
1. The bomb inherits the balloon's velocity at the moment of release.
2. Time of flight increases with the upward initial speed.

Answer: (b) 15, 20, 15

Explanation:
1. Time of flight of a ball = 2u/g = 4 s. Four balls spread over 4 s means a ball is thrown every 1 s.
2. The other three balls were thrown 1 s, 2 s and 3 s earlier.
3. \( h=ut-\frac12gt^2 \): t = 1 s gives 20 − 5 = 15 m; t = 2 s gives 40 − 20 = 20 m; t = 3 s gives 60 − 45 = 15 m.

Teacher's Note:
1. Balls are in the air for 4 s and thrown at regular intervals, so the interval is 1 s.
2. The middle ball is at the top of its path (maximum height 20 m).

Answer: (a) Velocity rises with slope increasing up to t1, then keeps rising with slope decreasing, becoming horizontal at t2 (S-shaped curve)

Explanation:
1. The slope of the v-t graph equals the acceleration.
2. From 0 to t1 acceleration increases from zero, so the slope increases gradually (curve bends upward).
3. From t1 to t2 acceleration is still positive but decreases to zero, so the velocity keeps increasing with decreasing slope and the graph becomes horizontal at t2.

Teacher's Note:
1. Acceleration is always positive here, so velocity never decreases.
2. A zero slope at t = 0 and at t = t2 gives the S-shaped curve.

Answer: (a) 50 m/s

Explanation:
1. The motion is symmetric about the highest point, so the time to reach the top is \( \frac{4+6}{2}=5\ \text{s} \).
2. \( u=gt=10\times5=50\ \text{m/s} \)

Teacher's Note:
1. The two times at which the body crosses a given height are symmetric about the time of the top.
2. At the top the velocity is zero, so u = g × (time to top).

Answer: (b) \( -2n\beta^2x^{-4n-1} \)

Explanation:
1. \( a=v\frac{dv}{dx} \).
2. \( \frac{dv}{dx}=-2n\beta x^{-2n-1} \).
3. \( a=\beta x^{-2n}\cdot(-2n\beta x^{-2n-1})=-2n\beta^2x^{-4n-1} \)

Teacher's Note:
1. When velocity is given as a function of position, use a = v dv/dx.
2. Add the powers of x when multiplying.

Answer: (c) \( \frac12f_0T \)

Explanation:
1. f = 0 when t = T.
2. \( v_x=\int_0^Tf_0\left(1-\frac tT\right)dt=f_0\left[t-\frac{t^2}{2T}\right]_0^T=f_0\left(T-\frac T2\right)=\frac12f_0T \)

Teacher's Note:
1. Velocity is the integral of acceleration with respect to time.
2. The area under the a-t graph (a triangle of base T and height f0) is (1/2)f0T.

Answer: (d) go on increasing with time

Explanation:
1. \( v=\frac{dx}{dt}=-a\alpha e^{-\alpha t}+b\beta e^{\beta t} \).
2. The acceleration \( \frac{dv}{dt}=a\alpha^2e^{-\alpha t}+b\beta^2e^{\beta t} \) is always positive, so the velocity keeps increasing with time.

Teacher's Note:
1. The sign of the acceleration decides whether velocity grows or falls.
2. Both terms of the acceleration are positive for positive constants.

Answer: (d) \( \frac{\alpha\beta t}{\alpha+\beta} \)

Explanation:
1. Let the maximum velocity be v, reached after time \( t_1=\frac v\alpha \); the retardation phase lasts \( t_2=\frac v\beta \).
2. \( t_1+t_2=t\Rightarrow v\left(\frac1\alpha+\frac1\beta\right)=t \).
3. \( v=\frac{\alpha\beta t}{\alpha+\beta} \)

Teacher's Note:
1. The maximum speed is reached at the instant the acceleration changes to retardation.
2. The sum of the two stage durations equals the total time.

Answer: (d) −4 m/s²

Explanation:
1. \( a_x=\frac{d^2x}{dt^2}=-4\ \text{m/s}^2 \) and \( a_y=\frac{d^2y}{dt^2}=0 \).
2. The acceleration is constant, so it is −4 m/s² at t = 2 s as well.

Teacher's Note:
1. Differentiate twice with respect to time for each coordinate separately.
2. Linear terms in t do not contribute to acceleration.

Answer: (a) A straight line, with a negative value at x = 0 that increases linearly to zero at x = x0

Explanation:
1. \( v=v_0\left(1-\frac{x}{x_0}\right) \).
2. \( a=v\frac{dv}{dx}=v_0\left(1-\frac x{x_0}\right)\left(-\frac{v_0}{x_0}\right)=-\frac{v_0^2}{x_0}\left(1-\frac x{x_0}\right) \).
3. So a is negative at x = 0 (retardation) and rises linearly to zero at x = x0.

Teacher's Note:
1. Use a = v dv/dx when velocity is given as a function of displacement.
2. The particle slows down, so the acceleration must be negative, and it vanishes when v = 0.

Answer: (a) \( \propto t^2 \)

Explanation:
1. \( \frac{dx}{dt}=\alpha\sqrt x\Rightarrow\int_0^x\frac{dx}{\sqrt x}=\alpha t \).
2. \( 2\sqrt x=\alpha t\Rightarrow x=\frac{\alpha^2t^2}{4} \), so x is proportional to t².

Teacher's Note:
1. Separate the variables and integrate both sides.
2. Constant acceleration α²/2 also gives x ∝ t².

Answer: (d) 0, 21.4

Explanation:
1. He starts and ends at O, so the net displacement is zero.
2. Distance covered: \( 1+\frac\pi2\times1+1=2+\frac\pi2\approx3.57\ \text{km} \).
3. Time = 10 min = 1/6 h, so \( v_{av}=\frac{3.57}{1/6}\approx21.4\ \text{km/h} \)

Teacher's Note:
1. Displacement depends only on the end points; speed uses the total path length.
2. Convert minutes to hours.

Answer: (c) 2.5 m/s²

Explanation:
1. Speed \( 72\ \text{km/h}=20\ \text{m/s} \). During the reaction time the car covers 20 m, leaving 80 m.
2. \( v^2=u^2-2as\Rightarrow0=400-2a(80)\Rightarrow a=2.5\ \text{m/s}^2 \)

Teacher's Note:
1. The car keeps moving at constant speed during the reaction time.
2. Convert km/h to m/s by multiplying by 5/18.

Answer: (a) 5 s

Explanation:
1. The packet starts with 12 m/s upward. Take up as positive: \( -65=12t-5t^2 \).
2. \( 5t^2-12t-65=0\Rightarrow t=\frac{12\pm\sqrt{144+1300}}{10}=\frac{12\pm38}{10} \).
3. The positive root is t = 5 s.

Teacher's Note:
1. The packet continues upward initially because it shares the balloon's velocity.
2. Discard the negative root.

Answer: (b) A closed circular loop

Explanation:
1. In one-dimensional motion the particle has exactly one velocity at any instant.
2. A closed loop (circle) gives two values of velocity at the same time, which is impossible.

Teacher's Note:
1. A graph that is not a function of time (two values at the same t) cannot describe a real motion.
2. The other curves are single-valued functions of t.

Answer: (a) A single straight line of negative slope crossing the time axis at the highest point

Explanation:
1. Velocity \( v=u-gt \) is a straight line with slope −g.
2. It is positive while rising, zero at the highest point and negative while falling, so the line crosses the t-axis once and continues downward.

Teacher's Note:
1. Speed would give a V-shape, but velocity keeps its sign.
2. The acceleration is constant at −g throughout.

Answer: (a) A velocity-time graph that is a straight line rising with a positive intercept

Explanation:
1. \( v=u+at \) is a straight line with slope a and intercept u.
2. (b) shows a changing acceleration; (c) shows zero acceleration; (d) shows constant velocity.

Teacher's Note:
1. Uniform acceleration means a straight-line v-t graph with non-zero slope.
2. Displacement-time for uniform acceleration is a parabola.