Question

s=1/3t³-7/2t²+12t+1
Find acceleration when velocity=0​

Answer 1

Answer:

Given,

s= 1/3t^3-7/2t^2 + 12t +1

v = ds/dt = 1/3*3t^2 - 7/2*2t + 12 +0

v = t^2 - 7t + 12

When v = 0

0 = t^2 - 7t + 12

0= t^2-3t-4t +12

0= t(t-3) - 4(t-3)

0 = (t-4)(t-3)

(t-4)=0 , (t-3)=0

t = 4s, t= 3s

Now,

a = dv/dt

a = 2t-7+0

a=2t-7

acceleration at t=4s

a4 = 2*4-7 = 8-7= 1m/s^2

acceleration at t = 3s

a3= 2*3-7 = 6-7= -1m/s^2

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