Question

Acceleration of a particle increases uniformly from $2m/s^2$ to $12m/s^2$ in 5 sec starting from rest. If particle starts from rest at t = 0, then

• A. Distance travelled in first 2 sec is 6 m
• B. Average velocity of first 4 sec is 9.33 m/s
• C. Average acceleration of first 2 sec is $4m/s^2$
• D. Displacement of first 2 sec is 6.67 m

Answer 1

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

The correct options are
B Average velocity of first 4 sec is 9.33 m/s
C Average acceleration of first 2 sec is $4m/s^2$
D Displacement of first 2 sec is 6.67 m

From the given data,
$a=2t+2$
${\int }_0^{v}dv={\int }_0^t(2t+2)dt$
$\Rightarrow v=t^2+2t$ -- (1)
${\int }_0^{x}dx={\int }_0^t(t^2+2t)dt$
$\Rightarrow x=\frac{t^3}{3}+t^2$ --(2)
$x\left(2\right)=\frac{2^3}{3}+2^2=6.66m$
$<v>=\frac{\Delta x}{\Delta t}=\frac{x\left(4\right)}{4}=9.33$ $\Rightarrow <a>=\frac{\Delta v}{\Delta t}=\frac{v\left(2\right)}{2}=4m/s^2$