Question

The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration -5.50 m/s2 for 4.45 s, making straight skid marks 61.1 m long ending at the tree. With what speed does the car then strike the tree? m/s

Answer 1

We must use the kinematical equation:

$s=ut+1/2at^{2}$ where u is the initial velocity of the car, a is the acceleration of the car and t is the time taken to cover the distance s.

We choose the coordinate axes such that the x-axis is taken to be along the direction of motion of the car with its positive sense in that direction.

The origin of the coordinate system is at the point where the driver applies the brakes.

Then substituting the values of s, a and t in the above equation we obtain:

$61.1=u*4.45-1/2*5.5*4.45^{2}$. From this we obtain $u=(61.1+1/2*5.5*4.45^{2})/4.45=26 m/s.$

This is the initial velocity of the car.

Now the final velocity of the car i.e when it strikes the tree can be obtained from the equation

$v=u+at=26-5.5*4.45=1.5 m/s.$

This is the speed with which the car strikes the tree.

The above problem is that of motion in one dimension with uniform acceleration so that the laws of rectilinear motion apply to it.