Question

A driver in a car traveling at a speed of 50 mi/h sees a deer 50 m away on the road. Calculate the minimum constant acceleration that is necessary for the car to stop without hitting the deer (assuming that the deer does not move in the meantime).

Answer 1

Answer:

a= - 0.79 m/s²

Explanation:

Given that

Speed ,u = 20 mi/h

We know that

1 mi/h= 0.44 m/s

Therefore ,u = 8.94 m/s

Distance ,s= 50 m

Lets take the acceleration of the car = a m/s²

The final speed of the car ,v = 0 m/s

We know that

v²= u² + 2 a s

Now by putting the values

0²= 8.94² + 2 x a x 50

$a=-\dfrac{8.94^2}{2\times 50}\ m/s^2$

a= - 0.79 m/s²

Therefore the acceleration will be - 0.79 m/s².