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/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

Answer 1

Answer:

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 $a_{avg}$ will be;

$a_{avg}$  = [$V_{c}$  - $V_{A}$] / Δt

$V_{c}$ is 17.4 ft/s,   $V_{A}$ is 13.2 ft/s and Δt is 3.00 s

so we substitute

$a_{avg}$  = [17.4 - 13.2] / 3

$a_{avg}$  = 4.2 / 3

$a_{avg}$  =  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;

$a_{avg}$  = Δv / Δt

now substitute [$V_{c}$  - $V_{B}$] for Δv and t' for Δt

$a_{avg}$  = [$V_{c}$  - $V_{B}$] / t'

$V_{B}$ = $V_{c}$ - $a_{avg}$( t' )

so we substitute 17.4 ft/s for $V_{c}$, 1.20 s for t' and  $a_{avg}$  =  1.4 ft/s²

$V_{B}$ = 17.4 - (1.4 × 1.20)

$V_{B}$ = 17.4 - 1.68

$V_{B}$ = 15.72 ft/s

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