A car moves along a circular track of radius 250 ft such that its seed for a short period of time 0<_ t <_ 4s, is v = 3 (t - t^2) ft/s, where t is in seconds.

a. Determine the magnitude of its acceleration when t = 3s.

b. How far has it traveled in t = 3s?

Question

Hannah Odom

Engineering

28 October, 02:16

A car moves along a circular track of radius 250 ft such that its seed for a short period of time 0<_ t <_ 4s, is v = 3 (t - t^2) ft/s, where t is in seconds.

a. Determine the magnitude of its acceleration when t = 3s.

b. How far has it traveled in t = 3s?

+1

Answers (1)

Know the Answer?

Averie · Answer 1

Answer: a) - 5 ft/s²

B) 4.5 ft

Explanation: Radius = 250ft

Velocity V = 3 (t-t²) ft/s

A). When t = 3 s, the acceleration is

dv/dt = 1-2t

dv/dt = 1 - 2 (3)

= 1-6

= - 5 ft/s²

B. How far it traveled in 3 sec

Distance = 3t²/2 - t³/3 ft

Substituting 3

Distance = 27/2 - 9

= 13.5 - 9

= 4.5 ft

A car moves along a circular track of radius 250 ft such that its seed for a short period of time 0<_ t <_ 4s, is v = 3 (t - t^2) ft/s, where t is in seconds.a. Determine the magnitude of its acceleration when t = 3s.b. How far has it traveled in t = 3s?

A car moves along a circular track of radius 250 ft such that its seed for a short period of time 0<_ t <_ 4s, is v = 3 (t - t^2) ft/s, where t is in seconds.

a. Determine the magnitude of its acceleration when t = 3s.

b. How far has it traveled in t = 3s?