JEE Physics Motion In A Straight Line MCQs Set 10

Answer: (b) \(\frac{3}{5}\)

Explanation:
1. Let the maximum speed be \(v\). The acceleration and retardation stages have average speed \(\frac{v}{2}\), so their times are \(\frac{2x}{v/2} = \frac{4x}{v}\) and \(\frac{6x}{v/2} = \frac{12x}{v}\).
2. The uniform stage takes \(\frac{4x}{v}\). Total time \(= \frac{20x}{v}\) and total distance \(= 12x\).
3. Average speed \(= \frac{12x}{20x/v} = \frac{3v}{5}\), so the ratio is \(\frac{3}{5}\).

Teacher's Note:
1. Average speed is total distance divided by total time.
2. On a straight uniform-acceleration stage from or to rest, the mean speed is half the maximum.

Answer: (b) \(25\ \text{m s}^{-1}\)

Explanation:
1. Let the maximum velocity be \(v\). The acceleration and retardation phases each last \(\frac{v}{5}\) s, so the total time is \(20 + \frac{v}{5}\).
2. Area under the graph \(= v\left(20 - \frac{v}{5}\right) + 2\cdot\frac{1}{2}\cdot v\cdot\frac{v}{5} = 20v\).
3. Average velocity \(= \frac{20v}{20 + v/5} = 20\), so \(v = 20 + \frac{v}{5}\), giving \(v = 25\ \text{m s}^{-1}\).

Teacher's Note:
1. The area under a v-t graph is the displacement.
2. Write the trapezium as a rectangle plus two triangles.

Answer: (b) \(20\ \text{m s}^{-1}\)

Explanation:
1. \(v = u - gt = 40 - 10\times 2\).
2. \(v = 20\ \text{m s}^{-1}\) (upward).

Teacher's Note:
1. Take upward as positive and \(g\) as negative.
2. The ball is still rising because it takes \(4\ \text{s}\) to reach the top.

Answer: (d) \(A\) feels that the body is going up, while \(B\) feels that it is going down

Explanation:
1. The body leaves the balloon with the balloon's upward velocity, so for \(A\) it initially moves up.
2. Relative to the balloon, the body starts with zero relative velocity and then accelerates downward relative to \(B\) (the balloon moves at constant velocity while the body slows).
3. So \(A\) sees it go up, and \(B\) sees it go down.

Teacher's Note:
1. Motion depends on the observer's frame.
2. A dropped body keeps the velocity of its carrier at the instant of release.

Answer: (c) \(2g\left(\frac{t_1 + t_2}{4}\right)^2\)

Explanation:
1. By symmetry, the time to reach the top is \(\frac{t_1 + t_2}{2}\).
2. The maximum height above the point of projection is \(H = \frac{1}{2}g\left(\frac{t_1 + t_2}{2}\right)^2\).
3. \(H = \frac{g(t_1 + t_2)^2}{8} = 2g\left(\frac{t_1 + t_2}{4}\right)^2\).

Teacher's Note:
1. The time up to the top is the average of the two times at a given height.
2. Free fall from the top takes the same time as the upward journey.

Answer: (a) a straight line through the origin

Explanation:
1. Both stones have the same acceleration \(g\), so the relative acceleration is zero.
2. The relative velocity stays equal to the initial velocity \(u\) of the second stone, so \(s = ut\).
3. The graph is a straight line through the origin.

Teacher's Note:
1. Equal accelerations cancel in the relative motion.
2. The separation grows uniformly.

Answer: (b) The total displacement of the particle is zero

Explanation:
1. From \(0\) to \(10\ \text{s}\) the area above the axis is \(\frac{1}{2}\times 10\times 10 = 50\ \text{m}\). From \(10\) to \(20\ \text{s}\) the area below is \(-50\ \text{m}\).
2. Displacement \(= 50 - 50 = 0\), and the total distance is \(100\ \text{m}\) at the end of \(20\ \text{s}\) (and \(50\ \text{m}\) at \(10\ \text{s}\)).
3. The slope is \(\frac{10}{5} = 2\ \text{m s}^{-2}\) at its maximum, so statements (a), (c), (d) are false.

Teacher's Note:
1. Areas below the axis count as negative displacement but positive distance.
2. The slope of a v-t graph is the acceleration.

Answer: (b) the acceleration may be zero at \(t = 2\ \text{s}\)

Explanation:
1. Zero velocity at one instant says nothing about the displacement before it, so (a) need not hold.
2. A particle at rest has zero velocity and zero acceleration, so the acceleration may be zero.
3. If the acceleration is non-zero, the velocity changes after \(t = 2\ \text{s}\), so (c) and (d) are not necessarily true.

Teacher's Note:
1. Velocity and acceleration are independent quantities at an instant.
2. At the top of a vertical throw, velocity is zero but acceleration is \(g\).

Answer: (d) \(1 : (\sqrt{2} - 1) : (\sqrt{3} - \sqrt{2})\)

Explanation:
1. Time to fall a height \(nh\) from rest is \(T_n = \sqrt{\frac{2nh}{g}}\), so \(T_1 : T_2 : T_3 = 1 : \sqrt{2} : \sqrt{3}\).
2. The time for each successive height is \(t_1 = T_1\), \(t_2 = T_2 - T_1\), \(t_3 = T_3 - T_2\).
3. So \(t_1 : t_2 : t_3 = 1 : (\sqrt{2} - 1) : (\sqrt{3} - \sqrt{2})\).

Teacher's Note:
1. The body speeds up, so later equal heights take less time.
2. Subtract cumulative times to get the time for each section.

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

Explanation:
1. Let the maximum velocity be \(v\), reached after \(t_1 = \frac{v}{\alpha}\), and the car stops after a further \(t_2 = \frac{v}{\beta}\).
2. \(t = t_1 + t_2 = v\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) = v\,\frac{\alpha + \beta}{\alpha\beta}\).
3. \(v = \frac{\alpha\beta}{\alpha + \beta}\,t\).

Teacher's Note:
1. The area under the v-t graph is not needed here, only the two times.
2. The result is the same as the combination formula for two resistances in parallel.

Answer: (d) \(C\)

Explanation:
1. The instantaneous velocity is the slope of the distance-time graph.
2. The slope is smallest near \(A\) (flat start) and \(D\) (flattening at the top).
3. The curve is steepest at the middle of the rise, at \(C\), so the velocity is maximum there.

Teacher's Note:
1. The slope of the tangent gives the instantaneous velocity.
2. The steepest point of an S-curve is its point of inflection.

Answer: (c) \(129.6\)

Explanation:
1. The retardation is the same in both cases, so \(s = \frac{v^2}{2a} \propto v^2\).
2. \(s_2 = 40\times\left(\frac{90}{50}\right)^2 = 40\times 3.24\).
3. \(s_2 = 129.6\ \text{m}\).

Teacher's Note:
1. Stopping distance varies as the square of the speed.
2. Both speeds are in the same units, so no conversion is needed in the ratio.

Answer: (b) The acceleration of the particle is \(2a\)

Explanation:
1. \(v = \frac{dx}{dt} = 4 + 2a(t - 2)\), and \(a_{particle} = \frac{dv}{dt} = 2a\).
2. At \(t = 0\), \(v = 4 - 4a\), which is not \(4\ \text{m s}^{-1}\) in general.
3. At \(t = 0\), \(x = -8 + 4a\), which is not zero in general. So only (b) is always true.

Teacher's Note:
1. Differentiate once for velocity and twice for acceleration.
2. The origin of time here is shifted by \(2\ \text{s}\).

Answer: (c) a line rising from zero at \(x = 0\) to a positive value at \(x = 100\), then a jump to a negative value and a line rising back to zero at \(x = 200\)

Explanation:
1. Use \(a = v\frac{dv}{dx}\). For \(0 < x < 100\), \(v = kx\), so \(a = k^2x\), which rises linearly from zero.
2. For \(100 < x < 200\), \(v = k(200 - x)\) and \(\frac{dv}{dx} = -k\), so \(a = -k^2(200 - x)\). It is negative and rises linearly to zero at \(x = 200\).
3. At \(x = 100\), \(a\) jumps from \(+100k^2\) to \(-100k^2\), which is graph (c).

Teacher's Note:
1. When velocity is given against displacement, use \(a = v\frac{dv}{dx}\).
2. Acceleration is positive while speeding up and negative while slowing down.

Answer: (b) \(v = u + \frac{1}{2}at^2\)

Explanation:
1. \(\frac{dv}{dt} = f = at\).
2. Integrating: \(v = u + \int_0^t at\,dt\).
3. \(v = u + \frac{1}{2}at^2\).

Teacher's Note:
1. The equations \(v = u + at\) hold only for constant acceleration.
2. For variable acceleration, integrate.

Answer: (c) \(A\) and \(B\) have equal average velocities from \(0\) to \(t_0\)

Explanation:
1. \(P\) is lower than \(Q\) on the x-axis, so \(OP < OQ\) and \(A\) lives closer.
2. \(A\)'s line starts at \(t = 0\) while \(B\)'s starts later, and the lines cross, so \(B\) overtakes \(A\).
3. Average velocity \(= \frac{\text{displacement}}{t_0}\), and the displacements \(OP\) and \(OQ\) differ, so the average velocities are not equal. Statement (c) is incorrect.

Teacher's Note:
1. The steeper line on an x-t graph means the greater speed.
2. Equal times but different displacements give different average velocities.

Answer: (d) \(g\) downward

Explanation:
1. Once the stone is released, the only force on it is gravity.
2. So its acceleration with respect to the ground is \(g\) downward, whatever the elevator is doing.
3. The elevator's acceleration affects only the relative acceleration.

Teacher's Note:
1. The acceleration of a freely falling object depends only on the forces acting on it.
2. The initial velocity equals the elevator's, but the acceleration does not.

Answer: (d) \(80\ \text{s}\)

Explanation:
1. The relative speed is \(20 - 7.5 = 12.5\ \text{m s}^{-1}\).
2. The relative distance to be covered is the sum of the lengths \(= 200 + 800 = 1000\ \text{m}\).
3. \(t = \frac{1000}{12.5} = 80\ \text{s}\).

Teacher's Note:
1. To overtake completely, the faster train must cover both lengths relative to the slower one.
2. Use the relative velocity of trains moving in the same direction.

Answer: (a) \(0.7\ \text{m s}^{-1}\)

Explanation:
1. The retardation is \(a = \frac{F}{m} = \frac{2.5\times 10^{-2}}{0.02} = 1.25\ \text{m s}^{-2}\).
2. \(v^2 = u^2 - 2as = 1 - 2(1.25)(0.2) = 0.5\).
3. \(v = \sqrt{0.5} \approx 0.7\ \text{m s}^{-1}\).

Teacher's Note:
1. Convert grams to kilograms and centimetres to metres.
2. Use \(v^2 = u^2 - 2as\) when time is not given.

Answer: (a) \(-9\ \text{m s}^{-1}\)

Explanation:
1. \(v = \frac{dS}{dt} = 3t^2 - 12t + 3\) and \(a = \frac{dv}{dt} = 6t - 12\).
2. The acceleration is zero at \(t = 2\ \text{s}\).
3. \(v = 3(4) - 12(2) + 3 = -9\ \text{m s}^{-1}\).

Teacher's Note:
1. Set the second derivative to zero to find the instant.
2. Then substitute that time in the first derivative.

Answer: (d) \(10\ \text{s}\)

Explanation:
1. Take north as positive. The parrot's velocity relative to the train is \(-5 - 10 = -15\ \text{m s}^{-1}\), so the relative speed is \(15\ \text{m s}^{-1}\).
2. The parrot must cover the length of the train relative to it.
3. \(t = \frac{150}{15} = 10\ \text{s}\).

Teacher's Note:
1. Opposite directions add their speeds in relative motion.
2. The length of the train is the relative distance.