Question

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

• A. $850\text{m}$
• B. $700\text{m}$
• C. $800\text{m}$
• D. $750\text{m}$

Answer 1

The correct option is D $750\text{m}$

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

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,
$v=u+at$
$\Rightarrow v=0+\left(2\right)\left(10\right)$
$\Rightarrow v=u_2=20\text{m/s}$
Now, $S_2=\left(20\right)\left(30\right)=600\text{m}$

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

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

Hence option D is the correct answer