Question

A car starts from rest at $t=0$ and for the first $4$ seconds of its rectilinear motion, the acceleration $a$ (${\text{in ms}}^{-2}$ ) at time $t$ (in seconds) after starting is given by $a=6–2t$. Choose the correct option(s).

• A. The maximum velocity of the car is $6\text{m/s}$
• B. The velocity of the car after $4$ seconds is $8\text{m/s}$
• C. The distance travelled up to $4$ seconds is $\frac{80}{3}\text{m}$
• D. The maximum velocity of the car is $9\text{m/s}$

Answer 1

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

The correct options are
B The velocity of the car after $4$ seconds is $8\text{m/s}$
C The distance travelled up to $4$ seconds is $\frac{80}{3}\text{m}$
D The maximum velocity of the car is $9\text{m/s}$

$a=6–2t$
$\Rightarrow \frac{dv}{dt}=6–2t$
For maximum velocity, $\frac{dv}{dt}=0$
$\Rightarrow 6-2t=0$
$\therefore t=3\text{s}$
$\frac{dv}{dt}=6-2t$
On integrating both the sides, we get
$\underset{0}{\overset{v}{\int }}dv=\underset{0}{\overset{t}{\int }}(6-2t)dt$
$v=6t-t^2.....\text{(i)}$
Maximum velocity $=6\left(3\right)-3^2=18-9=9\text{m/s}$
After $4$ seconds, $v=6t-t^2=6\times 4-16=24-16=8\text{m/s}$
From equ. $\text{(i)}$, we can write
$\frac{dx}{dt}=6t-t^2$
$\underset{0}{\overset{x}{\int }}dx=\underset{0}{\overset{t}{\int }}(6t-t^2)dt$
$\Rightarrow x=3t^2-\frac{t^3}{3}.....\text{(ii)}$
Putting $t=4\text{s}$, in equ. $\text{(ii)}$, we get
$x=\frac{80}{3}\text{m}$