Question

A particle starts from rest, accelerates at $2m/s^2$ for $10s$ and then goes with constant speed for $30s$ and then decelerates at $4m/s^2$ till it stops. What is the distance travelled by it?

• A. $750m$
• B. $800m$
• C. $700m$
• D. $850m$

Answer 1

The correct option is A $750m$

Applying $2^{nd}$ equation of motion for $1^{st}$ two parts
Part I: Constant acceleration
$S_1=u_{1}t+\frac{1}{2}{a}_1{t}^2$
$=0+\frac{1}{2}\left(2\right)(10{)}^2$
$=100m$

Part II: Uniform motion
$S_2=u_{2}t+\frac{1}{2}{a}_2{t}^2$
$=u_2\left(30\right)$
To find $u_2$, we will use $1^{st}$ equation of motion from the beginning to end of $1^{st}$ part,
$v=u+at$
$\Rightarrow v=0+\left(2\right)\left(10\right)$
$\Rightarrow v=u_2=20m/s$
Now, $S_2=\left(20\right)\left(30\right)=600m$

Applying $3^{rd}$ equation of motion for the third part,
$v_3^2-u_3^2=2a{S}_3$
Here, $v_3=0$ and $u_3=v_2=20m/s$
$\Rightarrow 0-(20{)}^2=2(-4\left)\right(S_3)$
$\Rightarrow -400=-8S_3$
$\Rightarrow S_3=50m$

$\therefore S_1+S_2+S_3=100+600+50=750m$