physics questions is there a chance of the car hitting?

Question

A driver in a car traveling at a speed of 49.7 units mi/h
sees a deer 105 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 in units of m/s^2 .

Answer 1

First, you need to convert mph to ms^-1

1 mph = 0.44704 msֿ¹
49.7 mph = 22.22 msֿ¹

So you must accelerate from 22.22 msֿ¹ to 0 msֿ¹ in 105m or less

Use the equations

v² = u² + 2as
v= final velocity
u = initial velocity
a = acceleration
s = displacement

0² = 22.22² + 2*a*105
-693.63 = 210a
-2.35 = a

You must accelerate at at least -2.35msֿ² (a decceleration of 2.35msֿ² or more)

Answer 2

I agree with Tom. Now go do the rest of your homework yourself.

Answer 3

t=105m/49.7m/s
t=2.11s
a=at^2
a=(V1-V2)/2
a=49.7-0)/2
a=24.85m/s^2
s=at^2/2
105m=24.85^2m/s^2/2
52.5m=24m/s2
52.5m/24.85m/s^2
2.11m/s