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Starting from rest the acceleration of a particle is 2(t-1). the velocity of the particle at t=5s is?

Starting from rest the acceleration of a particle is 2(t-1). the velocity of the particle at t=5s is?

Grade:11

6 Answers

Aditya
17 Points
6 years ago
Answer is 15 m/s .Integrate 2t-2 and apply limit from 0 to v for velocity since the body starts from rest and 0 to 5 for the time. You will get your answer
Ayushmaan Singh
32 Points
6 years ago
We know, A=dv/dtdv=AdtPutting the value of A in the above equation, and integrating `v` from 0 to v` since we require v` here.And integrating time from 0 to 5 s.Solving, we get: v`=t^2-
gavini haritha
27 Points
6 years ago
They have given acceleration,and asked to find velocity at 5s,then we should integrate acceleration then we get t^2.t=5 then v=5×5=25m/s
Aditya Rana
24 Points
6 years ago
As we know that change in velocity divided by change in time is the acceleration, we can writedv/dt = adv = a dtNow integrate both the sides. Limit the velocity from 0 to v. Limit the time from 0 to 5.dv = 2 (t-1) dtAfter integration:V = t^2 - 2t.Now using the limits, v = 5^2 - 2 (5)V = 25 - 10 = 15.Hence the answer is 15m/s.
adarsh jha
13 Points
6 years ago
acceleration is 2t-2.
by integration we get velocity = t2-2t
now thwe limit is from 0 to 5.
therefor,velocity=52-10=25-10  =15.
 
 
Monalisa Roy
19 Points
6 years ago
Integrate 2t-2 with respect to t , you will get it as ( t^2-2t ) then put the value of t and hence answer is 15.

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