Acceleration of a motorbike

Problem

Acceleration of a motorbike. A motorbike starts from rest and reaches $30\;\text{m/s}$ in $5\;\text{s}$. Calculate the average acceleration and the distance covered.

Solution

Average acceleration. This is the ratio between the change in speed and the time taken. The motorbike starts from rest ($v_0 = 0$):

$a = \frac{\Delta v}{\Delta t} = \frac{30 - 0}{5} = 6{,}0\;\mathrm{m/s^2}$

Distance covered. With constant acceleration starting from rest:

$x = \frac{1}{2}\,a\,t^2 = \frac{1}{2} \cdot 6{,}0 \cdot 5^2 = 75\;\mathrm{m}$

The same result is reached by calculating the area of the triangle under the line $v(t)$ (highlighted in the figure): $x = \tfrac{1}{2} \cdot \text{base} \cdot \text{height} = \tfrac{1}{2} \cdot 5 \cdot 30 = 75\;\text{m}$.

$\ev{a = 6{,}0\;\mathrm{m/s^2}, \quad x = 75\;\mathrm{m}}$

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Topics: Cinematica Concepts: Accelerazione · Moto uniformemente accelerato