Question

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/sft/s at point A and 17.6 ft/sft/s at point C. The cart takes 3.00 ss to go from point A to point C, and the cart takes 1.90 ss to go from point B to point C. What is the car's speed at point B?

Answer 1

Answer:

15.99 ft/s

Explanation:

From Newton's equation of motion, we have

v = u + at

v = Final speed

u = initial speed

a = acceleration

t = time

now

for the points A and C

v = 17.6 ft/s

u = 13.2 ft/s

t = 3 s

thus,

17.6 = 13.2 + a(3)

or

3a = 17.6 - 13.2

3a = 4.4

or

a = 1.467 m/s²

Thus,

For Points A and B

v = speed at B i.e v'

u = 13.2 ft/s

a = 1.467 ft/s²

t = 1.90 s

therefore,

v' = 13.2 + (1.467 × 1.90 )

v' = 13.2  + 2.7867

v' = 15.9867 ≈ 15.99 ft/s