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# A Car is Travelling Along the Road at 8 Ms-1. It Accelerates at 1 Ms-2 for a Distance of 18 M. How Fast is It Then Travelling ? - CBSE Class 9 - Science

ConceptRate of Change of Velocity

#### Question

A car is travelling along the road at 8 ms-1. It accelerates at 1 ms-2 for a distance of 18 m. How fast is it then travelling ?

#### Solution

We have to find the final velocity of the moving object. And we have the following information
Initial velocity, (u) = 8 m/s
Acceleration, (a) = 1 m/s2
Distance, (s) = 18 m
So applying 3rd equation of motion to calculate the final velocity,
ν = sqrt(u^2 + 2as)
Where,
(a) - Acceleration
(ν) - Final velocity
(u) - Initial velocity
(s) - Distance
Put the values in the above equation to get the value of final velocity,
ν = sqrt(64 + 36) m/s
= sqrt(100) m/s
= 10 m/s

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Solution A Car is Travelling Along the Road at 8 Ms-1. It Accelerates at 1 Ms-2 for a Distance of 18 M. How Fast is It Then Travelling ? Concept: Rate of Change of Velocity.
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