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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

Sagot :

nuhulawal20

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

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