Question

1. a cyclist accelerates from 0 m/s to 8 m/s in 3 seconds. what is his acceleration? is this acceleration higher than that of a car which accelerates from 0 to 30 m/s in 8 seconds? a roller - coaster car rapidly picks up speed as it rolls down a slope. as it starts down the slope, its speed is 4 m/s. but 3 seconds later, at the bottom of the slope, its speed is 22 m/s. what is its average acceleration?

Explanation:

Step1: Recall acceleration formula

The formula for average acceleration is $a=\frac{v - u}{t}$, where $v$ is the final - velocity, $u$ is the initial - velocity, and $t$ is the time interval.

Step2: Calculate cyclist's acceleration

For the cyclist, $u = 0\ m/s$, $v = 8\ m/s$, and $t = 3\ s$. Substitute these values into the acceleration formula: $a_{cyclist}=\frac{8 - 0}{3}=\frac{8}{3}\approx2.67\ m/s^{2}$.

Step3: Calculate car's acceleration

For the car, $u = 0\ m/s$, $v = 30\ m/s$, and $t = 8\ s$. Substitute into the formula: $a_{car}=\frac{30 - 0}{8}=\frac{30}{8}=3.75\ m/s^{2}$.

Step4: Analyze if cyclist's acceleration is higher

Since $\frac{8}{3}\approx2.67\ m/s^{2}$ and $3.75\ m/s^{2}$, the cyclist's acceleration is not higher than the car's acceleration.

Answer:

The cyclist's acceleration is $\frac{8}{3}\ m/s^{2}\approx2.67\ m/s^{2}$, and the cyclist's acceleration is not higher than that of the car.