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 4.50 m/s at point A and 5.00 m/s at point C. The cart takes 4.00 s to go from point A to point C, and the cart takes 1.60 s to go from point B to point C. What is the cart's speed at point B

Answer 1

Vi = 4.8 m/s

Step-by-step explanation:

First we need to find the magnitude of constant tangential acceleration. For that purpose we use the following formula between points A and C:

a = (Vf - Vi)/t

where,

a = constant tangential acceleration from A to C = ?

Vf = Final Velocity at C = 5 m/s

Vi = Initial Velocity at A = 4.5 m/s

t = time taken to move from A to C = 4 s

Therefore,

a = (5 m/s - 4.5 m/s)/4 s

a = 0.125 m/s²

Now, applying the same equation between points B and C:

a = (Vf - Vi)/t

where,

a = constant tangential acceleration from A to B = 0.125 m/s²

Vf = Final Velocity at C = 5 m/s

Vi = Initial Velocity at B = ?

t = time taken to move from B to C = 1.6 s

Therefore,

0.125 m/s² = (5 m/s - Vi)/1.6 s

Vi = 5 m/s - (0.125 m/s²)(1.6 s)

Vi = 4.8 m/s