Question

A cyclist is initially moving at 15 m/s. He covers 45 m in the next 4.8s. Find:
a) his acceleration
b) his speed after 4.8 s

Answer 1

Answer on Question 54935, Physics, Mechanics | Kinematics | Dynamics

Question:

A cyclist is initially moving at $15\,m/s$. He covers $45\,m$ in the next $4.8\,s$. Find:

a) his acceleration

b) his speed after $4.8\,s$

Solution:

a) We can find the acceleration of the cyclist from the kinematic equation:

$$s = v_0 t + \frac{1}{2} a t^2,$$

where, $s$ is the displacement, $v_0$ is the initial velocity, $t$ is the time and $a$ is the acceleration.

From this equation we obtain the acceleration of the cyclist:

$$a = \frac{2(s - v_0 t)}{t^2} = \frac{2(45\,m - 15\,\frac{m}{s} \cdot 4.8\,s)}{(4.8\,s)^2} = \frac{-54\,m}{23.04\,s^2} = -2.34\,\frac{m}{s^2}.$$

The sign minus means that the cyclist decelerate when he covers the distance of $45\,m$.

b) In order to find the speed of the cyclist after $4.8\,s$ we use another kinematic equation:

$$v = v_0 + a t = 15\,\frac{m}{s} + \left(-2.34\,\frac{m}{s^2}\right) \cdot 4.8\,s = 15\,\frac{m}{s} - 11.23\,\frac{m}{s} = 3.77\,\frac{m}{s}.$$

Answer:

a) $-2.34\,\frac{m}{s^2}$.

b) $3.77\,\frac{m}{s}$.

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