Question

if t = x^1/2 + 3, then find acceleration at t=2 sec?​

Answer 1

Explanation:

Position of the particle x=2t

2

−t

3

Velocity v=

dt

dx

=4t−3t

2

Acceleration a=

dt

dv

=4−6t

So, a( at t=2 s)=4−6×2=−8 m/s

2

Since, acceleration is depending on the time. SO, acceleration is not constant.

Hope it helps you

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