A cyclist increases their speed uniformly from $4.2 \, m/s$ to $6.3 \, m/s$ over a time interval of $5.3 \, s$. We need to calculate: a) The acceleration of the cyclist during this time. b) The distance the cyclist travels whilst overtaking. c) The average speed of the cyclist during this time.

1. Problem Description

A cyclist increases their speed uniformly from $4.2 \, m/s$ to $6.3 \, m/s$ over a time interval of $5.3 \, s$. We need to calculate:

a) The acceleration of the cyclist during this time.

b) The distance the cyclist travels whilst overtaking.

c) The average speed of the cyclist during this time.

2. Solution Steps

a) To find the acceleration, we use the formula:

$acceleration = \frac{change \, in \, velocity}{time}$

$a = \frac{v_f - v_i}{t}$

where $v_f$ is the final velocity, $v_i$ is the initial velocity, and $t$ is the time.

Given $v_i = 4.2 \, m/s$, $v_f = 6.3 \, m/s$, and $t = 5.3 \, s$.

$a = \frac{6.3 - 4.2}{5.3} = \frac{2.1}{5.3} \approx 0.396 \, m/s^2$

b) To find the distance traveled, we can use the following formula:

$d = v_i t + \frac{1}{2} a t^2$

Plugging in the values, we get:

$d = (4.2)(5.3) + \frac{1}{2}(0.396)(5.3)^2$

$d = 22.26 + \frac{1}{2}(0.396)(28.09)$

$d = 22.26 + 0.198(28.09)$

$d = 22.26 + 5.56182$

$d \approx 27.82 \, m$

Alternatively, we can use the equation:

$v_f^2 = v_i^2 + 2ad$

Rearranging for $d$:

$d = \frac{v_f^2 - v_i^2}{2a}$

$d = \frac{6.3^2 - 4.2^2}{2 \times 0.396} = \frac{39.69 - 17.64}{0.792} = \frac{22.05}{0.792} \approx 27.84 \, m$

c) To find the average speed, we can use the formula:

$average \, speed = \frac{total \, distance}{total \, time}$

Using the distance calculated earlier:

$average \, speed = \frac{27.82}{5.3} \approx 5.25 \, m/s$

Alternatively, since the acceleration is constant, the average speed is also the average of the initial and final speeds:

$average \, speed = \frac{v_i + v_f}{2} = \frac{4.2 + 6.3}{2} = \frac{10.5}{2} = 5.25 \, m/s$

3. Final Answer

a) The acceleration of the cyclist is approximately $0.396 \, m/s^2$.

b) The distance the cyclist travels whilst overtaking is approximately $27.8 \, m$.

c) The average speed of the cyclist during this time is $5.25 \, m/s$.