Question

A car is moving at 60km/hr when it begins to slow down with a deceleration of 1.50m/s^2.how long does it take to travel 70m as it slows down?

Answer 1

`t = 5.62 s`

Explanation:

Use the kinematic formula (from the set of 4 formulae)

`s = u*t + 1/2 a*t^2`

But first, it is necessary to get all the data in compatible units.

`60 cancel(km)/cancel(hr) * (1000 m)/(1 cancel(km)) * (1 cancel(hr))/(3600 s) = 16.67 m/s`

OK, plug the data into the formula

`70m = 16.67 m/s*t + 1/2*(-1.50m/s^2)*t^2`

Notice that I put a negative sign on that value for `a`. That is because deceleration is negative acceleration.

Rearrange into the format for the quadratic equation.

`-0.75 m/s^2*t^2 + 16.67 m/s*t - 70 m = 0`

Notice that if t is in seconds, the units of each term will be meters. Therefore we can drop the units. And note that I have multiplied thru by -1. The equation we will use the quadratic equation on is

`0.75 * t^2 - 16.67 * t + 70 = 0`

Using the quadratic equation to get the solution

`t = (16.67 +-sqrt(16.67^2 - 4*0.75 * (70)))/(2*0.75)`

`t = (16.67 +-sqrt(277.9 - 210))/1.5`

`t = (16.67 +-8.24)/1.5`

The 2 results for t are

`t = (16.67 +8.24)/1.5 = 16.6 s`
and
`t = (16.67 -8.24)/1.5 = 5.62 s`

The 5.62 s result is the correct one for this question. (The 16.6 s result is the time that would result when the deceleration continues after the car comes to a stop. That would be the 2nd time the car was at that 70 m point.)

I hope this helps,
Steve