# Please help me with this physics question and explain how to do it please. A 1423‐kg car is moving along a level highway with a speed of 26.4 m/s. The driver takes the foot off the accelerator...

Please help me with this physics question and explain how to do it please.

A 1423‐kg car is moving along a level highway with a speed of 26.4 m/s. The driver takes the foot off the accelerator and the car experiences a braking force of 901‐N over a distance of 106 m. Determine the speed of the car after traveling this distance.

*print*Print*list*Cite

### 1 Answer

Hello!

We know Newtons Second Law:

`F=m*a.`

This equation is a vector one, but here it is sufficient to consider its projection on a horizontal line. The gravity force and the normal force are vertical, so they are canceled out. The only force remaining is a braking force (horizontal), let it be denoted F. And its value is given along with the value of m.

So a=F/m is known and it is a constant. Consider that the braking force acting to the left and actually F and a is negative. Let's F and a be positive but remember that they act to the left (if car goes right, along the x-axis).

Then the distance travelled is

`d(t) = V_0*t-(a*t^2)/2`

and the velocity is

`V(t) = V_0-a*t.`

`V_0` is known (26.4 m/s), `a=901/1423 approx 0.63 (m/s^2).`

It is given that `d(t_1)=106m,` or `V_0*t_1-(a*t_1^2)/2=106.`

It is a quadratic equation for `t_1` , solve it:

`26.4*t_1 - (0.63)/2*t_1^2=106,`

or

`0.315*t_1^2 - 26.4*t + 106=0.`

`t_1=(13.2+-sqrt(13.2^2-106*0.315))/0.315.`

Select "-" sign, for "+" `V` will be negative.

`t_1 approx 4.2 (s).`

Therefore the velocity `V` will be `V_0-a*t_1 = 26.4-0.63*4.2 approx` **23.8 (m/s)**. This is the answer.