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 10.6 ft/s at point A and 15.6 ft/s at point C. The cart takes 4.00 s to go from point A to point C, and the cart takes 1.80 s to go from point B to point C. What is the cart's speed at point B

Answer 1

a)
`a_(avg)=1.25ft/s^2`

b)
`v_b=13.35ft/s`

Step-by-step explanation:

From the question we are told that:

Speed at point A
`v_A=10.6ft/s`

Speed at point C
`v_C=15.6ft/s`

Time from Point A to C
`T_(ac)=4.00s`

Time from Point B to C
`T_(bc)=1.80s`

Generally the equation for acceleration From A to B is mathematically given by

`a_(avg)=(v_c-v_a)/(\triangle t)`

`a_(avg)=(15.6-10.6)/(4.0 )`

`a_(avg)=1.25ft/s^2`

Generally the equation for cart speed at B is mathematically given by

`a_(avg)=(v_c-v_a)/(T_(bc))`

`v_b=v_c-a_(avg)*T_(bc)`

`v_b=15.6ft/s-(1.25ft/s^2)(1.80)`

`v_b=13.35ft/s`