Question

A bicycle accelerates uniformly from rest to a speed of 7 m/s over a distance of 35 meters. What was its acceleration?

Answer 1

Answer:

0.7 ms^-2

Explanation:

Initial velocity (u) = 0 m/s

Final velocity (v) = 7 m/s

Distance (s) = 35 m

Acceleration (a) = ?

We know,

${v}^{2} = {u}^{2} + 2as \\ a = \frac{ {v}^{2} - {u}^{2} }{2s} \\ a = \frac{ {7}^{2} - 0 }{2 \times 35} \\ a = \frac{49}{70} = \frac{7}{10} = 0.7 m {s}^{ - 2}$

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Answer 2

Given :

• Initial velocity, u = 0 m/s

• Final velocity, v = 7 m/s

• Distance, s = 35 m

To find :

• Acceleration, a

According to the question,

By using Newtons third equation of motion,

➞ v² = u² + 2as

Where,

• v = Final velocity
• u = Initial velocity
• a = Acceleration
• s = Distance

➞ Substituting the values,

➞ (7)² = (0)² + 2 × a × 35

➞ 49 = 0 + 70a

➞ 49 - 0 = 70a

➞ 49 = 70a

➞ 49 ÷ 70 = a

➞ 0.7 = a

So,the acceleration is 0.7 m/s².

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More formulas :

Newton's first equation of motion :

• v = u + at

Newton's second equation of motion :

• s = ut + ½ at²

Newton's third equation of motion :

• v² = u² + 2as