Question

A car increases its speed from 60km/hr to 65km/hr while a bicycle goes from rest to 3 m/s, which undergoes greater acceleration after both travels in 10 seconds?

Answer 1

Answer on Question 65757, Physics, Mechanics, Relativity

Question:

A car increases its speed from $60 \, \text{km/h}$ to $65 \, \text{km/h}$ while a bicycle goes from rest to $3 \, \text{m/s}$. Which undergoes greater acceleration after both travels in 10 seconds?

Solution:

Let's first convert $km/h$ to $m/s$:

$$v_{i \text{ car}} = 60 \, \frac{\text{km}}{\text{h}} \cdot \frac{1000 \, \text{m}}{1 \, \text{km}} \cdot \frac{1 \, \text{h}}{3600 \, \text{s}} = 16.6 \, \frac{\text{m}}{\text{s}},$$ $$v_{f \text{ car}} = 65 \, \frac{\text{km}}{\text{h}} \cdot \frac{1000 \, \text{m}}{1 \, \text{km}} \cdot \frac{1 \, \text{h}}{3600 \, \text{s}} = 18 \, \frac{\text{m}}{\text{s}}.$$

We can find the acceleration of the object from the kinematic equation:

$$v_f = v_i + a t,$$

here, $v_i$ is the initial speed of the object, $v_f$ is the final speed of the object, $a$ is the acceleration of the object and $t$ is the time.

Then, from this formula we can find the acceleration of the car and bicycle after both travels in 10 seconds:

$$a_{\text{car}} = \frac{v_f \text{ car} - v_i \text{ car}}{t} = \frac{18 \, \frac{\text{m}}{\text{s}} - 16.6 \, \frac{\text{m}}{\text{s}}}{10 \, \text{s}} = 0.14 \, \frac{\text{m}}{\text{s}^2},$$ $$a_{\text{bicycle}} = \frac{v_f \text{ bicycle} - v_i \text{ bicycle}}{t} = \frac{3 \, \frac{\text{m}}{\text{s}} - 0 \, \frac{\text{m}}{\text{s}}}{10 \, \text{s}} = 0.3 \, \frac{\text{m}}{\text{s}^2}.$$

As we can see, $a_{\text{bicycle}} > a_{\text{car}}$, so the bicycle undergoes greater acceleration after both travels in 10 seconds.

Answer:

The bicycle undergoes greater acceleration after both travels in 10 seconds.

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