Question

A car starting from rest accelerated at a rate of 0 .5 m/s up to 2 km what would be the final velocity of the car and how much time would it take to cover 1.6​​

Answer 1

Using third equation of kinematics

`\begin{gathered}\\ \rm\longmapsto v^2=u^2+2as\end{gathered}`

⟼v

2

=u

2

+2as

`\begin{gathered}\\ \rm\longmapsto v^2=0^2+2(0.5)(1.6)\end{gathered}`

⟼v

2

=0

2

+2(0.5)(1.6)

Answer 2

• Initial velocity=u=0m/s
• Acceleration=a=0.5m/s^2
• Distance=s=1.6m
• Final velocity=v=?

Using third equation of kinematics

`\\ \rm\longmapsto v^2=u^2+2as`

`\\ \rm\longmapsto v^2=0^2+2(0.5)(1.6)`

`\\ \rm\longmapsto v^2=1.6`

`\\ \rm\longmapsto v=√(1.6)`

`\\ \rm\longmapsto v=0.4m/s`

Now using 1st equation of motion

`\\ \rm\longmapsto v=u+at`

`\\ \rm\longmapsto t=v-u/a`

`\\ \rm\longmapsto t=(0.4-0)/(0.5)`

`\\ \rm\longmapsto t=(4)/(5)`

`\\ \rm\longmapsto t=0.8s`