Grade 11Mechanics

The velocity v of a body moving along a straight line varies with time t as v=2t^2*e^-t,where v is in m/s and t is in second . The acceleration of body is zero at t=?

Soumyadeep

11 Years agoGrade 11

2 Answers

AUREA

11 Years ago

v=2t²e^t

then

a=2te^t(t+2)

a=0 at t=0 or t=-2

since negative time has no physical significance, therefore t=0

nidhi mishra

7 Years ago

Given-

v=2t²e^-t

a=dv/dt =0= 2e^-t(dt²/dt) + 2t² (de^-t/dt)

0= 4t e^-t +2t²[ {de^-t/d(-t)}×{d(-t)/dt }]

0= 4te^-t - 2t²e^-t

0= 2te^-t(2-t)

Either 2te^-t=0

t=0

Or 2-t =0

t=2

So t =0 or t=2

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The velocity v of a body moving along a straight line varies with time t as v=2t^2*e^-t,where v is in m/s and t is in second . The acceleration of body is zero at t=?

Soumyadeep

2 Answers

AUREA

nidhi mishra

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