Question

A cyclist travelling at a uniform acceleration of 2.5m/s²passes through two points p and q in a straight line ,her speed at a point p is 20m/s and the distance between the points is 100m .calculate her speed at point q

Answer 1

Explanation:

Given

acceleration (a)= 2.5 m/s²

initial velocity (u)= 20 m/s

distance (s)= 100 m

final velocity (v) =?

By Newton's 3rd equation of motion

$v²-u²= 2as \\ \\ {v}^{2} - {20}^{2} = 2 \times 2.5 \times 100 \\ \\ {v}^{2} - 400 = 500 \\ {v}^{2} = 500 +400 \\ {v}^{2} = 900 \\ v = \sqrt{900} \\ v = 30 \: \: \frac{m}{s}$

Hence velocity be 30 m/s

Answer 2

Answer:

Velocity at 'q' (final velocity) = 30m/s

Explanation:

Given Information -

• Acceleration (a) = 2.5m/s²
• Distance (s) =100m
• Velocity at p (Initial velocity=u) = 20m/s

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To Find -

• Velocity at q (final velocity =v)

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Solution -

What to do?

To find the final velocity with the given information, we'll follow the formula of Position Velocity Relation and will replace the value with the formula to attain our required solution.

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Formula -

• $\underline{\boxed{\sf\purple{2as= v² - u² }}}$

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Answer -

[replacing values with formula]

$2(2.5 \times 100) = {v}^{2} - {u}^{2} \\ \\ 2( \frac{25}{10} \times 100) = {v}^{2} - {20}^{2} \\ \\ 2 \times 250 = {v}^{2} - 400 \\ \\ 500 + 400 = {v}^{2} \\ \\ 900 = {v}^{2} \\ \\ \sqrt{900 } = v \\ \\ 30 = v$

So, velocity at 'q' is 30m/s