Question

A roller coaster car is traveling at a constant 3 m/s when it reaches a downward slope. On the slope, the car accelerates at a constant rate of 4.5 m/s​ 2​ . The length of the slope is 45 meters. A) Find the velocity of the car at the bottom of the hill. B) Find the time of travel.

Answer 1

Velocity of the car at the bottom of the slope: approximately
`20.3\; \rm m \cdot s^(-2)`.

It would take approximately
`3.9\; \rm s` for the car to travel from the top of the slope to the bottom.

Step-by-step explanation:

The time of the travel needs to be found. Hence, make use of the SUVAT equation that does not include time.

• Let
  `v` denote the final velocity of the car.
• Let
  `u` denote the initial velocity of the car.
• Let
  `a` denote the acceleration of the car.
• Let
  `x` denote the distance that this car travelled.

`v^2 - u^2 = 2\, a\cdot x`.

Given:

• 
  `u = 3\; \rm m \cdot s^(-1)`.
• 
  `a = 4.5\; \rm m \cdot s^(-2)`.
• 
  `x = 45\; \rm m`.

Rearrange the equation
`v^2 - u^2 = 2\, a\cdot x` and solve for
`v`:

`\begin{aligned}v &= √(2\, a \cdot x + u^2) \\ &= \sqrt{2 * 4.5\; \rm m \cdot s^(-2) * 45\; \rm m + \left(3\; \rm m \cdot s^(-1)\right)^(2)} \\ &\approx 20.3\; \rm m \cdot s^(-1)\end{aligned}`.

Calculate the time required for reaching this speed from
`u = 3\; \rm m \cdot s^(-1)` at
`a = 4.5\; \rm m \cdot s^(-2)`:

`\begin{aligned}t &= (v - u)/(a) \\ &\approx (20.3\; \rm m \cdot s^(-1) - 3\; \rm m \cdot s^(-1))/(4.5\; \rm m \cdot s^(-2)) \approx 3.9\; \rm m \cdot s^(-1)\end{aligned}`.