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A particle starts from rest accelerates at. 2 m/secgor10seconds and then goes with constant speed for 30sc and then decelerates at 4 m/sec^2 till it stops .what is the distance travelled by it

A particle starts from rest accelerates at. 2 m/secgor10seconds and then goes with constant speed for 30sc and then decelerates at 4 m/sec^2 till it stops .what is the distance travelled by it

Grade:12th pass

1 Answers

URVASHI
24 Points
6 years ago
  1. 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 .
  2. Given that we have constant speed implies initial veocity is 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 .
  3. 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 = 750m

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