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. How much distance is between you and the deer when you come to a stop? What is the maximum speed you could have and still not hit the deer?

Question

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.   How much distance is between you and the deer when you come to a stop? What is the maximum speed you could have and still not hit the deer?

Serita Dewitt · Accepted Answer

Step 1 The acceleration of the body is defined as the rate of change of velocity and velocity as the rate of change of displacement. Step 2 Here Initial velocity, u=20ms Final velocity, v=0 a=−10ms2 Initial distance, d=35m Time travelling with constant speed is given by, u=dt d=u× =20ms×0.5s =10m Step 3 Noting that it involves deceleration, we write v=u+at 0=20ms+(−10ms2)× (10ms2)t=20ms t=20ms10ms2 =2 sec⁡onds Let us find the Distance covered during t=2s, s=ut+12at2 =(20ms×2s)+12×(−10ms2)×(2s)2 =40m−20m =20m Total distance=Time travelling with constant speed+Distance with deceleration =10m+20m =30m So total distance covered is 30m out of initial distance which is given 35m Distance covered between you and deer=Initial distance−total distance =35m−30m =5m Step 4 To find maximum speed Maximum speed is denoted as Vmax We will use kinematics equation, vf∈al2=vmax2+2ad2 Here vf∈al=0 Time travelling with constant speed is given by, vmax=d1t d1=vmax× =vmax×0.50s Distance left is, d2=35m−d1 =35m−d1, =35m−(vmax×0.50s) =35m−0.50Vmax Step 5

Melissa Moore · Accepted Answer

Answer: Explanation: Discount the time here; it&#039;s not important. It doesn&#039;t tell you how long it takes the car to stop, it only refers to reaction time, which means nothing in the scheme of things. The useful info is as follows: initial velocity =20ms final velocity =0ms a=−10ms/s and we are looking for the displacement. Use the following equation: v2=v02+2a△x where v is the final velocity, v0 is the initial velocity, a is the deceleration (since it&#039;s negative), and △x is displacement. Filling in: 02=(20)2+2(−10)△x and 0=400−20△x and −400=−20△x so △=20 meters

alenahelenash · Accepted Answer

Step 1 Firstly, we should calculate the distance traveled during the reaction time. Notice the car during that time is traveling with constant speed. v=dt Then d=v d1=20×0.5=10m (1) Step 2 Secondly, we should calculate the distance required to make the car stops when applying the breaks. vf2=vi2+2ad2 0=202+2(−10)d2 Notice we put negative in front of a because it is a declaration. d2=−202−20=20m (2) Step 3 From (1) and (2) The total distance traveled before the car stops is d1+d2=30m So the distance between the car and the deer when it stops: 35−30=5m Recall that the distance when the car driver noticed the deer was 35m (Given). Step 4 To calculate the maximum speed the car should have stiil not hit the deer. We have vf=0 and total distance traveled before stopping is 35m. The acceleration for the car is -10m/s. So we shall use the kinematic equation vf2=vi2+2ad but we must calculate the distance for reaction time which is with constant speed: d=vmaxt=0.5vmax Then the distance left is: d2=35−d1=35−0.5vmax vf2=vmax2+2ad2=vmax2+(2×(−10)×(35−12vmax)) 0=vmax2−20(35−12vmax) vmax2=700−10vmax vmax2+10vmax−700=0 It is a quadratic equation so to solve it we will use the mathematics law for quadratic. vmax=−b±b2−4ac2a Wherease a, b and c are the factors in front of variables: a=1 b=10 c=−700 Then vmax=−10±102−(4×−700)2=21.9,−31.9ms We will reject the negative solution. Therefore vmax=21.9ms

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