Question

Velocity variations of an object moving along a straight line are recorded in the enclosed figure.

• A. The acceleration is maximum in first two seconds
• B. The acceleration magnitude is maximum during the ninth second
• C. The object is farthest from the starting point at $16\text{s}$
• D. The displacement is largest at $7\text{s}$

Answer 1

Answer: (A), (D)

The correct options are
A The acceleration is maximum in first two seconds
D The displacement is largest at $7\text{s}$


7 line segments are there in this graph, out of which 3 are parallel to axis with slope 0.
Slopes of remaining line segments are $+2,-\frac{4}{3},-\frac{3}{2},1$ respectively.
Since slope of $v-t$ graph represents acceleration therefore, for maximum acceleration, option, (a) is correct.
Magnitude of acceleration at $(t=9\text{s})=\frac{3}{2}$
$\therefore$ Option (b) is incorrect.
From $t=0$ to $t=7\text{s}$, velocity is positive so, displacement will be maximum at $t=7\text{s}$.
Since displacement is maximum at $t=7\text{s}$, therefore object won't be farthest from starting point at 16 s.
Hence option (c) is incorrect.
After $t=7\text{s}$, velocity becomes zero and further after $t=8.5\text{s}$, displacement will start decreasing due to motion in the opposite direction.
So, option (d) is also correct.