The position of an object changes with time according to x(t) = 2t^3 + 5t + 1. Find
a)the velocity at times t = 0s, t = 2s and t = 4s. Find the acceleration at these instants
Answers
Answered by
3
Answer:
velocity = distance/time and
acceleration = velocity/time
x(t) = 2t³+5t+1
by differentiating dx/dt = 6t² + 5
dx/dt = velocity
v at t =o s = 6(0)² + 5 = 5m/sec
v at t = 2 s = 6(2)² + 5 = 29 m/sec
v at t = 4 sec = 6(4)² + 5 = 101 m/sec
a = d²x/dt² of 6t² + 5 [ 2nd derivative ]
a = 12t
a at t = 0 s = 12(0) = 0
a at t = 2 s = 12(2) = 24 m/sec²
a at t = 4 s = 12(4) = 48 m/sec²
Similar questions