The cart travels the track again and now experiences a constant tangential acceleration from point A to point C. The speeds of the cart are 13.2 ft/s at point A and 17.4 ft/s at point C. The cart takes 3.00 s to go from point A to point C, and the cart takes 1.20 s to go from point B to point C. What is the cart's speed at point B

the cart's speed at point B is 15.72 ft/s

Explanation:

 Given the data in the question;

The car travels from point A to C in 3.00 s, its average acceleration will be;

 = [  - ] / Δt

is 17.4 ft/s,   is 13.2 ft/s and Δt is 3.00 s

so we substitute

 = [17.4 - 13.2] / 3

 = 4.2 / 3

 =  1.4 ft/s²

so average acceleration of the cart between the points  A and B is 1.4 ft/s²

The instantaneous value of the velocity of the cart at point B will be;

 = Δv / Δt

now substitute [  - ] for Δv and t' for Δt

 = [  - ] / t'

= - ( t' )

so we substitute 17.4 ft/s for , 1.20 s for t' and    =  1.4 ft/s²

= 17.4 - (1.4 × 1.20)

= 17.4 - 1.68

= 15.72 ft/s

Therefore, the cart's speed at point B is 15.72 ft/s