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 3.80 m/s at point A and 6.10 m/s at point C . The cart takes 4.50 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

We can use the following kinematic equation to solve for the acceleration of the cart:

v_f = v_i + a*t

where v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and t is the time.

Using this equation for the motion from point A to point C, we have:

6.10 m/s = 3.80 m/s + a*(4.50 s)

Solving for a, we get:

a = (6.10 m/s - 3.80 m/s) / (4.50 s) = 0.71 m/s^2

Now we can use the same equation for the motion from point B to point C, with v_i = 3.80 m/s, a = 0.71 m/s^2, and t = 1.60 s, to solve for v_f:

v_f = v_i + a*t = 3.80 m/s + 0.71 m/s^2 * 1.60 s = 4.92 m/s

Therefore, the speed of the cart at point B is 4.92 m/s.