Question

calculate instantaneous acceleration of object at the time 5s , when the distance is given by x = t^3 + 3t^2 + 2t + 1 .

please help me ​

Answer 1

Solution:

s = dx/dt

Distance w.r.t. time

s = d (t³ + 3t² + 2t + 1)/dt

s = 3t² + 6t + 2

Now,

v = ds/dt

Velocity w.r.t. time

v = d (3t² + 6t + 2)/dt

v = 6t + 6

Acceleration w.r.t. time

a = dv/dt

a = d (6t + 6)/dt

a = 6 m/s²

Therefore, instantaneous acceleration is 6 m/s²