Question

At time $t=0$, a car moving along a straight line has a velocity of $16m{s}^{-1}$. It slow 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 $59m$
• C. The distance travelled by the particle in $10s$ is $94m$
• D. The velocity at $t=10s$ is $9m{s}^{-1}$

Answer 1

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

The correct options are
A The direction of velocity changes at $t=8s$.
B The distance travelled in $4s$ is approximately $59m$
C The distance travelled by the particle in $10s$ is $94m$

$A)$ $a=-0.5t$

$V=V_0+{\int }_0^{t}adt$

$=V_0+{\int }_0^t-0.5tdt$

$V=16-\frac{{0.5t}^2}{2}\to \left(1\right)$

A)$V=0$

$\Rightarrow 0=16-\frac{0.5}{2}{t}^2$

$32=0.5t^2$

$\Rightarrow t=8s$

Velocity changes its direction at $t=8s$.

$B)$ $x={\int }_0^{t}Vdt$

$\therefore x={\int }_0^4{[(16-\frac{0.5t^2}{2})dt=16t-\frac{0.5t^3}{6}]}_0^4$

$=58.677m$

$\approx 59m$

$C)$ $x={\int }_0^{10}Vdt$

$={\int }_0^{8}Vdt+\left|{\int }_8^{10}Vdt\right|$

$={[16t-\frac{0.5t^3}{6}]}_0^8+\left|{[16-\frac{0.5t^3}{6}]}_8^{10}\right|$

$=\frac{256}{3}+|-\frac{26}{3}|$

$x=94m$

$D)$ $V=V_0-\frac{0.5+2}{2}$ (from eqn$\left(1\right)$)

$\therefore$ $t=10s$,

$V=16-\frac{0.5\times {10}^2}{2}=-9m/s$.

Hence : Option $A,B$ & $C$