You’re driving down the highway late one night at 20 m/s

when a deer steps onto the road 35 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 10 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?

Short Answer

Expert verified

a. Before coming to a stop, the total distance of the car from the deer is, 5m.

b. The maximum speed of the car could have without hitting the deer is,22.36m/s.26.46m/s.

Step by step solution

01

Part a Step 1: Given information

Initial speed of the car, u=20m/s

Reaction time before striking the break, t=0.50s.

Acceleration of the car, role="math" localid="1648219673493" a=-10m/s2.

Distance between the car and the deer,d=35m.

02

Calculating the distance travelled and time consumed during deacceleration

The distance travelled before pressing the break is,

s=uts=20×0.50ms=10m

Time duration of deacceleration of the car during stepping on the brakes can be determines as,

v=u+at'0=20m/s-10m/s2×t't'=2010st'=2s

03

Calculating total distance travelled

Distance covered during the deacceleration,

S=ut'+12at'2S=20×2+12-10×22S=40-20mS=20m

Distance travelled during the uniform motion and deacceleration,(s+S)=10+20m=30m.

Before coming to a stop, the total distance from the deer is,

35-30m=5m.

04

Part b; Step 1: Given data

Acceleration of the car, a=-10m/s2.

Distance between the car and the deer, role="math" localid="1648219786110" d=-35m.

We have to find the maximum speed of the car without hitting the deer as vmax.

05

Determining the maximum speed

The total distance covered by the car from the starting time of its journey to stop without hitting the deer is,

d'=S+5md'=20+5md'=25m

The maximum speed of the car should have, can be found as, vmax2=vfinal2-2ad'

Now, vfinal=0

Therefore,

localid="1650273158159" vmax2=-2×10×25vmax2=500vmax=500vmax=22.36m/s

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