Question

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

A
750 m
B
800 m
C
700 m
D
850 m

Solution

## The correct option is A $$750\ m$$A)Initially particle starts from rest. So$$u=0 \ and \ a=2m/s^2 and t=10 sec$$ and formula for this case is: $$s=ut+(1/2)a(t^2)$$ and using this $$s = 0*10 + (1/2)*2*(10^2) = 100m$$  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 $$t=30 sec$$ (given). And from part $$u=20m/sec$$Now to get the value of distance in this part we use this formula :$$s=ut+(1/2)a(t^2)$$ and using this we get:$$20*30+(1/2)*0*(30^2) = 600m .$$Now $$a=(-)4m/sec^2$$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: $$0-20^2 = 2*(-4)*s => s =400/8 = 50m.$$ So using all the part info we get total distance travelled: 100+600+50 = 750Physics

Suggest Corrections

0

Similar questions
View More

People also searched for
View More