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 14.0 ft/s at point A and 19.0 ft/s at point C. The cart takes 3.50 s to go from point A to point C, and the cart takes 1.40 s to go from point B to point C. What is the cart's speed at point B

Answer 1

The cart's speed at point B is 21 ft/s.

Step-by-step explanation:

To find the speed of the cart at point B, we can use the equation

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.

Given that the cart takes 1.40 s to go from point B to point C and the final velocity at point C is 19.0 ft/s, we can rearrange the equation to solve for the acceleration:

a = (v - u) / t

Substituting the values, we get:

a = (19.0 ft/s - 14.0 ft/s) / 1.40 s = 5.0 ft/s²

Since the cart maintains a constant tangential acceleration, this acceleration is also the acceleration experienced from point A to point B.

To find the speed at point B, we can use the same equation with the given initial velocity of 14.0 ft/s and the acceleration of 5.0 ft/s²:

v = 14.0 ft/s + (5.0 ft/s²)(1.40 s)

Solving this equation gives:

v = 21 ft/s

Answer 2

`v_b` = 17.98 m / s

Step-by-step explanation:

To solve this exercise we will use the kinematics relations in one dimension, let's start by looking for the acceleration between points A and C

v = v₀ + a t

a =
`(v - v_o)/(t)`

a =
`(19 - 14)/(3.50)`

a = 1.43 m / s²

this acceleration is uniform throughout the path.

Let's find the velocity at point B

`v_c = v_b + a t`

`v_b = v_c - a t`

let's calculate

`v_b` = 19 - 1.43 1.40

`v_b` = 17.98 m / s