Question

Learning Goal: To practice Problem-Solving Strategy 2.1 Motion with constant acceleration You are 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 deceleration of your car is 10 m/s? . How much distance is between you and the deer when you come to a stop?

A.Sort the variable for the Period from t1=0s to t2=0.5 s based on the whether their values are know or unknown( ax,t,x1,xf,vf,v1)

Answer 1

During Reaction Time :

t1= 0 to t2 = 0.5 s

brakes are no applied so a = 0

x1 = vt = 20 ( 0.5 - 0) = 10 m

during reaction time 10m is covered.

During Braking:

a = -10 m/s2

Initial velocity = 20 m/s

it finally stops,

so final velocity v = 0

using, v^2 - u^2 = 2ad

0 - 20^2 = 2 x -10 x X2

X2 = 20 m

Total distance traveled before coming to stop = x1 + X2

= 10 + 20 = 30 m

deer was 35m front of car.

so distance remaining = 35-30 = 5 m .............Ans

Answer 2

The velocity is given by, Here, s is the displacement, t is the time Substitute 20 m/s for v and 0.50 s for t in above equation s-vt The actual distance d between the car and the deer at the point of hitting the brakes is, Here, D is the total distance between the deer and the man initially, s is the distance travelled in the reaction time and d is the is the actual distance between the deer and the man at the instant when the brakes are applied. Substitute 35 m for D and 10 m for x. d 35-10 - 25 m Therefore, the actual distance between the car and the deer is 25 m 2nd equation of motion is v2as Her v is the final velocity and u is the initial velocity Write the above equation in terms of s, y2-2 2a Substitute 0 for v, 20 m/s for u and -10 m/s for a in above equation. (0) -(20 m/s) 2(10 m s = 20 m The distance between the deer and the man when the man comes to stop is Substitute 25 m for d and 20 m for s in above equation. S 25-20