Question

A car accelerates from 5.0 m/s to 21 m/s at a constant rate of 3.0 m/s2, How far does it travel while accelerating?

Answer 1

`"69 m"` (rounded up to 2 significant figures)

Explanation:

Use equation of motion

`S = (v^2 - u^2) / (2a)`

Where

• `S =` Displacement covered when it acquires velocity `v`
• `u =` Initial velocity of car
• `v =` Velocity of car when it covers displacement `S`
• `a =` Acceleration of car

`S = (("21 m/s")^2 - ("5.0 m/s")^2) / ("2 × 3.0 m/s"^2) = "69 m"`

Answer 2

The distance is `=69.3m`

Explanation:

Apply the equation of motion

`v^2=u^2+2as`

The initial velocity is `u=5ms^-1`

The final velocity is `v=21ms^-1`

The acceleration is `a=3ms^-2`

The distance is

`s=(v^2-u^2)/(2a)`

`=(21^2-5^2)/(2*3)`

`=(26xx16)/(6)`

`=69.3m`