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 12.4 ft/s at point A and 19.4 ft/s at point C. The cart takes 4.00 s to go from point A to point C, and the cart takes 1.50 s to go from point B to point C.
What is the cart's speed at point B?

Answer 1

Answer:

 v_{B} = 15.03 ft / s

Explanation:

For this exercise let's use the accelerated motion relationships in one dimension

         v = vo + a t

         a = (v- v₀) /a

         a = (19.4 -12.4) /4.00

         a = 1.75 ft / s²

As we have the acceleration let's look for the speed for the time of 1.50 s

         $v_{B}$ = vo + a t

         v_{B} = 12.4 + 1.75 1.50

          v_{B} = 15.03 ft / s