Question

1. A body has a constant acceleration of 2m/s2 starting from rest.
How far has it travelled by the time that its velocity has reached 40m/s?

Answer 1

Answer:

By third equation of motion i.e:

v^2= u^2+ 2as

So,

40^2= 0^2+ 2×2×S

1600= 0 + 4S

therefore,

S= 1600/4

i.e 400m

your answer is 400m

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Answer 2

Given :

• Acceleration (a) = 2 m/s²
• Initial velocity (u) = 0 m/s
• Final velocity (v) = 40 m/s

To Find :

• Distance travelled when it reaches to speed of 40 m/s

Formula Used :

• v² - u² = 2as

Explanation :

We are given, constant acceleration is acting on the body and the magnitude of acceleration is 2 m/s², also initial velocity is given as 0 m/s as the body starts from the rest, final velocity is given as 40 m/s. We have to find out distance travelled by the body when the velocity is reached to 40 m/s .

So, we will use here Kinematics Equations, Use 3rd equation of Kinematics for solving this question.

$\bigstar \: \: \boxed{\sf{v^2 \: - \: u^2 \: = \: 2as}}$

Where,

• v is final velocity
• u is initial velocity
• a is acceleration of the body
• and s is distance

Calculation :

$\implies \sf{v^2 \: - \: u^2 \: = \: 2as} \\ \\ \implies \sf{40^2 \: - \: 0^2 \: = \: 2 \: \times \: 2 \: \times \: s} \\ \\ \implies \sf{1600 \: - \: 0 \: = \: 4s} \\ \\ \implies \sf{4s \: = \: 1600} \\ \\ \implies \sf{s \: = \: \dfrac{1600}{4}} \\ \\ \implies \sf{s \: = \: 400}$

$\therefore$ Distance travelled by the body when velocity reaches to 40 m/s is 400 m