Question

At time $t=0$, a car moving along a straight line has a velocity of $16m{s}^{-1}.$ It slows down with an acceleration of $-0.5tm{s}^{-2}$ where $t$ is in seconds. Mark the correct statement(s).

• A. The direction of velocity changes at $t=8s$
• B. The distance travelled in 4s is approximately 58.67 m
• C. The distance travelled by the particle in 10 s is 94 m
• D. The speed of particle at $t=10$ s is $9m{s}^{-1}$

Answer 1

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

The correct options are
A The direction of velocity changes at $t=8s$
B The distance travelled in 4s is approximately 58.67 m
C The distance travelled by the particle in 10 s is 94 m
D The speed of particle at $t=10$ s is $9m{s}^{-1}$

Given, $\frac{dv}{dt}=-0.5t$.
Integrate to get, $v-16=-0.25t^2$
Direction changes after v becomes negative, which occurs at $0-16=-0.25t^2$ or, $t=\sqrt{16\times 4}=8seconds.$
speed at 10 seconds is $v=16-0.25\times 100=-9$. Thus speed is magnitude, which is 9 m/s.
Integrate the velocity equation once more: $\frac{ds}{dt}=16-0.25t^2$ or, $s-s_0=16t-\frac{1}{12}{t}^3$ Assume $s_0=0$ for distance calculations (because its relative to some initial point)
In 4 seconds, distance = $16\times 4-\frac{1}{12}\times 64=64-\frac{16}{3}$ which is nearly 58.67m.
In 10 seconds, distance is sum of distance till 8 seconds and then 9th and 10th second (because velocity becomes negative, so displacement starts decreasing, thus we need to break it down)
= $16\times 8-\frac{1}{12}\times 512+(-16\times 10+\frac{1}{12}\times 1000+s_8)$
= $85.33+(-76.67+85.33)=94m$