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 [tex]a_{avg}[/tex] will be;
[tex]a_{avg}[/tex] = [[tex]V_{c}[/tex] - [tex]V_{A}[/tex]] / Δt
[tex]V_{c}[/tex] is 17.4 ft/s, [tex]V_{A}[/tex] is 13.2 ft/s and Δt is 3.00 s
so we substitute
[tex]a_{avg}[/tex] = [17.4 - 13.2] / 3
[tex]a_{avg}[/tex] = 4.2 / 3
[tex]a_{avg}[/tex] = 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;
[tex]a_{avg}[/tex] = Δv / Δt
now substitute [[tex]V_{c}[/tex] - [tex]V_{B}[/tex]] for Δv and t' for Δt
[tex]a_{avg}[/tex] = [[tex]V_{c}[/tex] - [tex]V_{B}[/tex]] / t'
[tex]V_{B}[/tex] = [tex]V_{c}[/tex] - [tex]a_{avg}[/tex]( t' )
so we substitute 17.4 ft/s for [tex]V_{c}[/tex], 1.20 s for t' and [tex]a_{avg}[/tex] = 1.4 ft/s²
[tex]V_{B}[/tex] = 17.4 - (1.4 × 1.20)
[tex]V_{B}[/tex] = 17.4 - 1.68
[tex]V_{B}[/tex] = 15.72 ft/s
Therefore, the cart's speed at point B is 15.72 ft/s
