Question

A particle starts from rest, accelerates at $2m/s^2$ for $10s$ and then goes for 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 $750m$

A)
Initially particle starts from rest. So$u=0anda=2m/s^{2}andt=10sec$ and formula for this case is:
$s=ut+\left(1/2\right)a\left(t^2\right)$ and using this
$s=0\ast 10+\left(1/2\right)\ast 2\ast \left({10}^2\right)=100m$
and using the given information and this formula

v = u + at

we get the value of final velocity and on substituting all values we get:

0+2*10 = 20m/sec
Given that we have constant speed implies initial velocity is the same as the final velocity.

v=u and using this information and the formula $v=u+at$

we get a=0 as t can’ t be zero. and $t=30sec$ (given). And from part $u=20m/sec$

Now to get the value of distance in this part we use this formula :

$s=ut+\left(1/2\right)a\left(t^2\right)$ and using this we get:

$20\ast 30+\left(1/2\right)\ast 0\ast \left({30}^2\right)=600m.$

Now $a=(-)4m/se{c}^2$

as particle is decelerating and given statement is it decelerates till is stops implies that final velocity is zero. And from above part $u=20m/sec$ and to get the value of distance we use this formula:

$v^2-u^2=2as$ and using this we get:

$0-{20}^2=2\ast (-4)\ast s=>s=400/8=50m.$

So using all the part info we get total distance travelled: 100+600+50 = 750