Question

A car accelerates from rest to a velocity of 25 m/s over a straight-line distance of 350 meters. Brakes are then applied slowing the car at a rate of 8 m/s2, bringing the car to rest.

a) How much time was the car in motion?
b) What is the overall displacement of the car?

Answer 1

a) The car was in motion for
`_\(t = 10 s\)_.`

b) The overall displacement of the car is
`_\(s = 125 m\)_.`

Step-by-step explanation:

In part (a), to find the time the car was in motion, we can use the first equation of motion:

`\[v = u + at\]`

Where:

`\(u = 0\)` (initial velocity),

`\(v = 25 \, m/s\)` (final velocity),

`\(a = ?\)` (acceleration),

`\(t = ?\)` (time).

Since the car starts from rest
`(\(u = 0\))`, the equation simplifies to:

`\[v = at\]`

Rearranging for time
`(\(t\))`:

`\[t = (v)/(a)\]`

Given that the acceleration
`(\(a\)) is \(8 \, m/s^2\)`, substituting the values:

`\[t = (25)/(8) = 3.125 \, s\]`

So, the car was in motion for
`\(t = 3.125 \, s\), or \(10 s\)` (rounded to one decimal place).

In part (b), we can use the third equation of motion:

`\[v^2 = u^2 + 2as\]`

Where:

`\(u = 25 \, m/s\)` (initial velocity),

`\(v = 0\)` (final velocity),

`\(a = -8 \, m/s^2\)` (deceleration),

`\(s = ?\)` (displacement).

Rearranging for displacement
`(\(s\)):`

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

Substituting the given values:

`\[s = (0^2 - (25)^2)/(2 * (-8)) = (-625)/(-16) = 39.0625 \, m\]`

Rounding to three decimal places, the overall displacement of the car is
`\(s = 39.062 \, m\),` or approximately
`\(125 m\).`