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 10ms2.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?

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You&#039;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 10ms2.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?

Bobby Hodges · Accepted Answer

Step 1Given dataOur driving speed is u=20ms.The distance between the car and the deer is db=35mThe reaction time interval is t=0.50s .The deceleration rate of car is a=−10ms2 .a) The distance traveled by car in the reaction time as,PSKu = frac{d}{t}PSKd_{1} = utaskd1=20m×0.5sd1=10mStep 2The distance required to stop the car due to deceleration as,v2=u2+2ad20=(20ms)2+2⋅(−10ms2)⋅d2d2=40020d2=20mThe total distance traveled by car before car stops is;d2=d1+d2d1=10m+20md1=30mTherefore, the distance between car and deer after car stop is calculated as,d=db−dtd=35m−30md=5mThus, the distance between the car and the deer is 5 m.

Janessa Bradshaw · Accepted Answer

(b)distance traveled before applying break is d1=10mDistance traveled when break is applied,d2=35−10=25mLet v be the maximum speed of the car, before applyng break,Then,0=v2=2ass=Distan⁡cetrave≤ddur∈gbreak∈g=25m0=v2−2×10×25⇒v=22.36ms

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