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# A car acclerates from rest at constant rate 2 ms-2 for sometime.Then retards at a constant rate of 4 ms-2 and comes to rest. it remains in motion for 3 seconds. then total distance covered by it is 7 years ago

first of all let time for acceleration be t seconds

so the time of decceleration is 3-t sec.

now,use v=u+at with u=0,a=2,t=tsec,

get v=2t...

now use v=u+at with u=2t v=0,a=-4 t=3-t

get t=2sec

now calculate distance with s=ut+1/2at^2 wit t=2 and t=1 sec respectively

7 years ago

in the easiest method first try to draw the velocity time graph and then calculate the displacement as area under the graph.

lets say that it has accelerated for t seconds then in order to come to rest it must decelerate for t/2 seconds at double rate.

that is it has accelerated for 2 seconds and then decelerated for 1 second.

as we know that the acceleration is the slope of the velocity time graph so the maximum velocity attained is clearly 4m/s.

so the graph obtained is a triangle with base of 3 units and altitude 4 units.

be clear about the graph and try to draw it yourself understanding clearly the logic behind each step.

so the total displacement is equal to (1/2)*3*4 = 6 m.

note that before it comes to stop the particle travels in one direction only.

so in this case the distance covered equals the displacement.

had the velocity become negative that is the particle would have decelerated a little more then clearly the displacement would have been less but the distance would have been more than the answer obtained.

this is because in that case the particle would have turned back and travelled a little in the opposite direction.