Constant acceleration and maximum deceleration

  • Thread starter andyman21
  • Start date
  • #1
13
0
You are driving down the highway late one night at 20 \rm 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/s^2 . How much distance is between you and the deer when you come to a stop?

I have already found the answer to be 5 meters between me and the deer and now it wants me to "Determine the time required for you to stop once you hit the brakes. Im confused on how to get this value. Any help would be appreciated. Thanks
 

Answers and Replies

  • #2
13
0
I completely forgot to put how i figured out my answer.... sorry for that .....

Vi = 20 m/s
Vf = 0.0 m/s
d = 35 m
Tr = 0.50 s
a = -10 m/s^2 --> Note this must be negative, as acceleration is "acting against" your line of travel. if it isn't, your math won't work.

First calculate how far you travel before you hit the brakes. This is easy, 6th grade math. Distance, rate, time.

D = R * T
D = Vi * Tr
D = (20 m/s) * (0.50 s)
D = 10 m

Second, how far you went after you hit the brakes.

Dd = [ Vf^2 - Vi^2 ] / [ 2 * (-10 m/s^2) ]
Dd = [ (0.0 m/s)^2 - (20 m/s)^2 ] / (-20 m/s^2)
Dd = [ -400 m^2/s^2 ] / (-20 m/s^2)
Dd = 20 m

Total distance traveled

Dt = D + Dd
Dt = (10 m) + (20 m)
Dt = 30 m

Does he survive?

Ds = d - Dt
Ds = (35 m) - (30 m)
Ds = 5 m


So you come to a stop 5 meters before hitting him
 
  • #3
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,167
507
yes. Now for the time it takes to stop after hitting the brakes.. use another of the motion equations.
 
  • #4
13
0
Would i plug my answer of 5m into say d=vi+vf/2xt ( so it would be 5=20+0/2xt?)
 
  • #5
PhanthomJay
Science Advisor
Homework Helper
Gold Member
7,167
507
Why would you use d =5? You stopped in 20 m. You can use that figure in your equation. Or simply use the motion equation relating v with acceleration and time as a check.
 

Related Threads on Constant acceleration and maximum deceleration

Replies
7
Views
2K
Replies
3
Views
11K
Replies
2
Views
12K
  • Last Post
Replies
13
Views
6K
Replies
3
Views
2K
  • Last Post
Replies
2
Views
2K
Replies
2
Views
3K
  • Last Post
Replies
1
Views
688
Top