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

To calculate the minimum constant acceleration required for the car to stop without hitting the deer, we can use the equation vf^2 = vi^2 + 2ad, where vf is the final velocity (0 m/s, since the car needs to stop), vi is the initial velocity (50 mi/h converted to m/s), a is the acceleration, and d is the distance (50 m). By substituting the values into the equation and solving for a, we can find the minimum constant acceleration required to stop the car without hitting the deer.

Step-by-step explanation:

To calculate the minimum constant acceleration required for the car to stop without hitting the deer, we can use the equation:

vf^2 = vi^2 + 2ad

where vf is the final velocity (0 m/s, since the car needs to stop), vi is the initial velocity (50 mi/h converted to m/s), a is the acceleration, and d is the distance (50 m).

By substituting the values into the equation and solving for a, we can find the minimum constant acceleration required to stop the car without hitting the deer.

The conversion from miles per hour (mi/h) to meters per second (m/s) can be done using the conversion factor 1 mi/h = 0.44704 m/s.

Therefore, the minimum constant acceleration required for the car to stop without hitting the deer is approximately -4.46 m/s^2 (negative value indicates deceleration).

Answer 2

a= - 0.79 m/s²

Step-by-step 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=-(8.94^2)/(2* 50)\ m/s^2`

a= - 0.79 m/s²

Therefore the acceleration will be - 0.79 m/s².