Question

A stone is projected vertically upwards at t = 0 second. The net displacement of stone is zero in time interval between t = 0 second to t = T seconds. Pick up the incorrect statement.

• A. From time $t=\frac{T}{4}$ second to $t=\frac{3T}{4}$ second, the average velocity is zero.
• B. The change in velocity from time t=0 to $t=\frac{T}{4}$ second is same as change in velocity from $t=\frac{T}{8}$ second to $t=\frac{3T}{8}$ second.
• C. The distance travelled from t = 0 to $t = \frac{T}{4}$ second is larger than distance travelled from $t = \frac{T}{4}$ second to $t = \frac{3T}{4}$ second.
• D. The distance travelled from $t = \frac{T}{2}$ second to $t = \frac{3T}{4}$ second half the distance travelled from $t = \frac{T}{2}$ second to t = T second.

Answer 1

Answer: (D)

(d)

Answer 2

The problem describes the motion of a stone projected vertically upwards. The net displacement of the stone is zero in the time interval $t=0$ to $t=T$ seconds, which means the stone returns to its initial position at $t=T$. This implies that the time taken to reach the maximum height is $T/2$. Let $u$ be the initial velocity and $g$ be the acceleration due to gravity.

Key Relations for Vertical Motion:

1. Time of flight (T): For zero net displacement, $0 = uT - \frac{1}{2}gT^2 \implies u = \frac{1}{2}gT$.
2. Time to reach maximum height ($t_{max}$): At maximum height, velocity $v=0$. So, $0 = u - gt_{max} \implies t_{max} = \frac{u}{g} = \frac{(1/2)gT}{g} = \frac{T}{2}$.
3. Maximum height (H): $H = ut_{max} - \frac{1}{2}gt_{max}^2 = u\frac{T}{2} - \frac{1}{2}g(\frac{T}{2})^2 = (\frac{1}{2}gT)\frac{T}{2} - \frac{1}{8}gT^2 = \frac{1}{4}gT^2 - \frac{1}{8}gT^2 = \frac{1}{8}gT^2$.
4. Position at time t: $y(t) = ut - \frac{1}{2}gt^2$.

Let's evaluate each statement:

Statement (a): From time $t=\frac{T}{4}$ second to $t=\frac{3T}{4}$ second, the average velocity is zero.

Average velocity = $\frac{\text{Net displacement}}{\text{Time interval}}$.

Let's find the position of the stone at $t=T/4$ and $t=3T/4$.

$y(T/4) = u(T/4) - \frac{1}{2}g(T/4)^2 = (\frac{1}{2}gT)\frac{T}{4} - \frac{1}{2}g\frac{T^2}{16} = \frac{1}{8}gT^2 - \frac{1}{32}gT^2 = \frac{3}{32}gT^2$.

$y(3T/4) = u(3T/4) - \frac{1}{2}g(3T/4)^2 = (\frac{1}{2}gT)\frac{3T}{4} - \frac{1}{2}g\frac{9T^2}{16} = \frac{3}{8}gT^2 - \frac{9}{32}gT^2 = \frac{12-9}{32}gT^2 = \frac{3}{32}gT^2$.

Since $y(T/4) = y(3T/4)$, the net displacement between these two times is zero.

Therefore, the average velocity is zero. Statement (a) is correct.

Statement (b): The change in velocity from time t=0 to $t=\frac{T}{4}$ second is same as change in velocity from $t=\frac{T}{8}$ second to $t=\frac{3T}{8}$ second.

The velocity of the stone is given by $v(t) = u - gt$.

The change in velocity over a time interval $[t_1, t_2]$ is $\Delta v = v(t_2) - v(t_1) = (u - gt_2) - (u - gt_1) = -g(t_2 - t_1)$.

This shows that the change in velocity depends only on the time interval, not on the specific start and end times, provided $g$ is constant.

For the first interval ($t=0$ to $t=T/4$): $\Delta v_1 = -g(T/4 - 0) = -\frac{gT}{4}$.

For the second interval ($t=T/8$ to $t=3T/8$): $\Delta v_2 = -g(3T/8 - T/8) = -g(\frac{2T}{8}) = -\frac{gT}{4}$.

Since $\Delta v_1 = \Delta v_2$, statement (b) is correct.

Statement (c): The distance travelled from t = 0 to $t = \frac{T}{4}$ second is larger than distance travelled from $t = \frac{T}{4}$ second to $t = \frac{3T}{4}$ second.

Let's use the positions calculated earlier:

$y(0) = 0$

$y(T/4) = \frac{3}{32}gT^2$

$y(T/2) = H = \frac{1}{8}gT^2 = \frac{4}{32}gT^2$

$y(3T/4) = \frac{3}{32}gT^2$

Distance travelled from $t=0$ to $t=T/4$:

During this interval, the stone is moving upwards.

$d_1 = |y(T/4) - y(0)| = |\frac{3}{32}gT^2 - 0| = \frac{3}{32}gT^2$.

Distance travelled from $t=T/4$ to $t=3T/4$:

In this interval, the stone moves upwards from $t=T/4$ to $t=T/2$ (maximum height), then downwards from $t=T/2$ to $t=3T/4$.

$d_2 = |y(T/2) - y(T/4)| + |y(3T/4) - y(T/2)|$

$d_2 = |\frac{4}{32}gT^2 - \frac{3}{32}gT^2| + |\frac{3}{32}gT^2 - \frac{4}{32}gT^2|$

$d_2 = \frac{1}{32}gT^2 + \frac{1}{32}gT^2 = \frac{2}{32}gT^2 = \frac{1}{16}gT^2$.

Comparing $d_1$ and $d_2$: $\frac{3}{32}gT^2$ vs $\frac{1}{16}gT^2 = \frac{2}{32}gT^2$.

Since $\frac{3}{32}gT^2 > \frac{2}{32}gT^2$, $d_1$ is indeed larger than $d_2$. Statement (c) is correct.

Statement (d): The distance travelled from $t = \frac{T}{2}$ second to $t = \frac{3T}{4}$ second half the distance travelled from $t = \frac{T}{2}$ second to t = T second.

Distance travelled from $t=T/2$ to $t=3T/4$:

This is the downward motion from max height.

$d_3 = |y(3T/4) - y(T/2)| = |\frac{3}{32}gT^2 - \frac{4}{32}gT^2| = |-\frac{1}{32}gT^2| = \frac{1}{32}gT^2$.

Distance travelled from $t=T/2$ to $t=T$:

This is the total downward motion from max height to the initial position.

$d_4 = |y(T) - y(T/2)| = |0 - \frac{4}{32}gT^2| = \frac{4}{32}gT^2 = \frac{1}{8}gT^2$.

Now, let's check if $d_3 = \frac{1}{2}d_4$:

Is $\frac{1}{32}gT^2 = \frac{1}{2} \left(\frac{1}{8}gT^2\right)$?

Is $\frac{1}{32}gT^2 = \frac{1}{16}gT^2$?

No, $\frac{1}{32}gT^2 \neq \frac{2}{32}gT^2$.

Therefore, statement (d) is incorrect.

The question asks for the incorrect statement.

The final answer is $\boxed{\text{(d)}}$

Solution Summary:

1. Establish key parameters: For a stone projected vertically upwards returning to its initial position at time T, the initial velocity is $u = gT/2$ and the maximum height is $H = gT^2/8$, reached at $t=T/2$.
2. Calculate positions: Using $y(t) = ut - \frac{1}{2}gt^2$, find positions at $t=0, T/4, T/2, 3T/4, T$.
   $y(0)=0$, $y(T/4)=\frac{3}{32}gT^2$, $y(T/2)=\frac{4}{32}gT^2$, $y(3T/4)=\frac{3}{32}gT^2$, $y(T)=0$.
3. Evaluate (a): Average velocity from $t=T/4$ to $t=3T/4$. Displacement is $y(3T/4) - y(T/4) = 0$. Average velocity is $0$. Correct.
4. Evaluate (b): Change in velocity. Change in velocity is $\Delta v = -g \Delta t$. For both intervals, $\Delta t = T/4$. So, change in velocity is the same. Correct.
5. Evaluate (c): Distance travelled.
   Distance from $t=0$ to $t=T/4$: $d_1 = |y(T/4)-y(0)| = \frac{3}{32}gT^2$.
   Distance from $t=T/4$ to $t=3T/4$: $d_2 = |y(T/2)-y(T/4)| + |y(3T/4)-y(T/2)| = \frac{1}{32}gT^2 + \frac{1}{32}gT^2 = \frac{2}{32}gT^2$.
   Since $\frac{3}{32}gT^2 > \frac{2}{32}gT^2$, $d_1 > d_2$. Correct.
6. Evaluate (d): Distance travelled.
   Distance from $t=T/2$ to $t=3T/4$: $d_3 = |y(3T/4)-y(T/2)| = \frac{1}{32}gT^2$.
   Distance from $t=T/2$ to $t=T$: $d_4 = |y(T)-y(T/2)| = \frac{4}{32}gT^2$.
   Is $d_3 = \frac{1}{2}d_4$? Is $\frac{1}{32}gT^2 = \frac{1}{2} \left(\frac{4}{32}gT^2\right)$? Is $\frac{1}{32}gT^2 = \frac{2}{32}gT^2$? No. Incorrect.

The incorrect statement is (d).