Question

a bike accelerates uniformly from rest to 10m/s over a distance of 30m . Determine the acceleration of the bike

Answer 1

Answer:

(5/3) m/s²

Explanation:

Use equation of motion

• v² - u² = 2aS

a = (v² - u²)/(2S)

= [(10 m/s)² - (0 m/s)²] / (2 × 30 m)

= (100/60) m/s²

= (5/3) m/s²

Answer 2

$\maltese \: \red{\mid{\underline{\overline{\textbf{Answer}}}\mid}}$

Acceleration = 5/3 m/s^2

$\maltese \: \red{\mid{\underline{\overline{\textbf{Explanation}}}\mid}}$

According to third equation of motion , the square of final velocity of a body is equals to sum of its initial velocity and double of product of its acceleration and distance. This is represented by :

$\maltese \: {v}^{2} = {u}^{2} + 2as$

In this equation :

• v = final velocity
• u = initial velocity
• a = acceleration
• s = distance

In this question :-

• Initial velocity = 0 since it starts from rest.
• Final velocity = 10m/s
• Distance = 30m
• Acceleration = ?

By third equation of motion we can say :-

$\maltese \: a = \frac{ {v}^{2} - {u}^{2} }{2s}$

After inputting the values we get :

$\maltese \: a = \frac{ {10}^{2} - {0}^{2} }{2 \times 30}$

$\maltese \: a = \frac{100}{60}$

$\maltese \: a = \cancel \frac{100}{60} = \frac{5}{3}$

Therefore acceleration of bike is 5/3m/s^2.

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BY SCIVIBHANSHU

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