Question

A point moves in a straight line such that its distance s from the start in time t is equal to
s=1/4 t^4-4t^3+16t^2

a)At what time its velocity equal to zero?
b)At what time was the point at it's starting point?​

Answer 1

Explanation:

hope this will help.

First differentiate position S wrt dt .

Put value of V equal to 0 then find factor of t .

Answer 2

Answer:

Explanation:

a) At what times was the point at its starting position?

b) At what times its velocity equal to zero?

solution : (a) we have to find time at s = 0

so, s = 1/4 t⁴ - 4t³ + 16t² = 0

or, t² [ 1/4 t² - 4t + 16] = 0

or, (t² - 16t + 64)/4 = 0

or, t² - 16t + 64 = 0

or, t² - 2(8)t + (8)² = 0

or, (t - 8)² = 0

or, t = 8sec

at t = 8sec was the point at its starting position.

(b) we have to find time at v = 0

we know, velocity is the rate of change of displacement with respect to time.

i.e., v = ds/dt

⇒v = d(1/4 t⁴ - 4t³ + 16t²)/dt

= 1/4 d(t⁴)/dt - 4 d(t³)/dt + 16 d(t²)/dt

= 1/4 × 4t³ - 4 (3t²) + 16(2t)

= t³ - 12t² + 32t

now, v = t³ - 12t² + 32t = 0

or, t(t² - 12t + 32) = 0

or, t(t² - 8t - 4t + 32) = 0

or, t(t - 8)(t - 4) = 0

or, t = 0, 4, 8

hence, velocity will be zero at t = 0s, 4s and 8s .

                                                                                                                                           

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