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

Answer:

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.

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 $v$:

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

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

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