Question

The figure shows the velocity-time graph for a train as it moves along a straight line. At time $t=0$ the train passes a point $A$ and is moving at $3\text{m/s}$. The farthest point from $A$ reached by the train in the $120\text{s}$ period is $P$. Choose the correct option(s)

• A. The value of $t$ at the instant when the train reaches $P$ is $40\text{s}$.
• B. The magnitude of the acceleration of the train in the time interval $50<t<80$ ($t$ in seconds) is $\frac{1}{15}{\text{m/s}}^2$
• C. The distance of the train from $A$ at the end of the $120\text{s}$ period is $35\text{m}$.
• D. The distance of the train from $A$ at the end of the $120\text{s}$ period is $105\text{m}$.

Answer 1

Answer: (A), (B), (C)

The correct options are
A The value of $t$ at the instant when the train reaches $P$ is $40\text{s}$.
B The magnitude of the acceleration of the train in the time interval $50<t<80$ ($t$ in seconds) is $\frac{1}{15}{\text{m/s}}^2$
C The distance of the train from $A$ at the end of the $120\text{s}$ period is $35\text{m}$.

From the graph, we can clearly see that the particle reverses its direction after $t=50\text{s}$. From $t=40\text{s}$ to $t=50\text{s}$, the particle is at rest, so the particle is at the farthest point at $t=40\text{s}$.
Acceleration in this interval $50<t<80$ ($t$ in seconds) is
$|a|=\frac{\Delta v}{\Delta t}=\left|\frac{-2}{30}\right|=\frac{1}{15}{\text{m/s}}^2$
Area under curve from $t=0$ to $40\text{s}$
$=30\times 3+\frac{1}{2}\times 10\times 3=90+15=105\text{m}$.
Area from $t=50\text{s}$ to $120\text{s}$. $=\frac{1}{2}\times 70\times -2=-70\text{m}$.
We know that area under the $v-t$ graph represents displacement
$\therefore$ Total displacement of the particle from $A=[105–70]=35\text{m}$.