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A particle starts from rest accelarates 2m/s2 for 10 s and then goes at constant speed for 30 s and tgen deaccelarates at 4m/s2 till it stops. What is distance covered by it

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3 years ago

Amarjit
105 Points
```							First of all.Applying s=ut+1/2at² we get s=100mSecondly we calculate the final velocity by that time by applying the formula v=u+at. v comes to be 20m/s.Now since now there is movement with constant speed hence now applying d=s×t we get d=600m.Now since there is deacceleration so the final velocity will be 0 so using the formula v=u-at we get t=5 sec for the body to stop.Now again applying the formula s=ut-1/2at² we get the distance to be 50m.Hence finally the total distance covered by it will be 750m.
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3 years ago
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