You're driving down the highway late one night at 17 m/s when a deer steps onto the road 45 m in front of you. Your reaction time before stepping on the brakes is 0.50 s, and the maximum deceleration of your car is 12 m/s2.
a. How much distance is between you and the deer when you come to a stop?
b. What is the maximum speed you could have and still not hit the deer?

A)33m

B)32.86m/s

Step-by-step explanation:

Using the equation of motion

V^2=U^2-2aS

Where V=final velocity of the car=0

U=initial velocity=17m/s

a=deceleration=-12m/s^2

S= distance covered by car before stopping

Substituting into the equation

0=17^2-2×12×S

0=289-24S

24S=289

S=289/24

S=12m.

The distance between the car and the deer=45-12=33m

B) Using same equation

Distance,S is now 45m,which is the distance between the car and the deer.while deceleration remains same

V^2=U^2-2aS

0=U^2-2×12×45

U^2=1080

U=√1080

U=32.86m/s

A. So therefore distance between deer and car after stopping= 36.5 - 12.04 = 24.46m

B. u = 29.6m/s = initial velocity

Step-by-step explanation:

Given the reaction time as 0.5s

So the distance traveled before applying breaks = u*rt

Where u = initial velocity = 17m/s

rt = reaction time = 0.5s

So distance traveled = 0.5 * 17 = 8.5m

So distance between deer and car before applying breaks = 45 - 8.5 = 36.5m

So therefore using kinematics formula to solve for distance covers after applying breaks

V² = u² + 2as

Where v = final velocity = 0

u = initial velocity = 17m/s

a = deceleration = -12m/s²

s = distance covered

Solving for s = u²/2a

s = 17²/(2*12) = 12.04m

So therefore distance between deer and car after stopping= 36.5 - 12.04 = 24.46m

B. Using the same kinematic formula to derive maximum initial velocity attainable to cover distance of 36.5m before stopping

V² = u² + 2as

V= 0

a = deceleration = -12m/ s

s = distance covered = 36.5m

u = (2as)^½

u = ( 2* 12 * 36.5)^½

u = 29.6m/s