Constant acceleration and maximum deceleration

In summary, when driving at 20 m/s, with a reaction time of 0.50 s and a maximum deceleration of 10 m/s^2, the distance between you and a deer when you come to a stop is 5 meters. To determine the time it takes to stop after hitting the brakes, you can use the motion equation d = vi + vf/2 * t, where d is the distance traveled, vi is the initial velocity, vf is the final velocity (which is 0 in this case), and t is the time. Alternatively, you can use the motion equation v = vi + a * t, where v is the final velocity, vi is the initial velocity, a is the acceleration
  • #1
andyman21
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. I am confused on how to get this value. Any help would be appreciated. Thanks
 
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  • #2
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 mSo you come to a stop 5 meters before hitting him
 
  • #3
yes. Now for the time it takes to stop after hitting the brakes.. use another of the motion equations.
 
  • #4
Would i plug my answer of 5m into say d=vi+vf/2xt ( so it would be 5=20+0/2xt?)
 
  • #5
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 to Constant acceleration and maximum deceleration

1. What is constant acceleration?

Constant acceleration refers to a situation where an object is moving with a uniform increase or decrease in velocity. This means that the object's speed changes by the same amount over equal time intervals.

2. How is constant acceleration calculated?

Constant acceleration is calculated using the equation a = (vf - vi)/t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time interval. This equation is known as the acceleration formula.

3. What is maximum deceleration?

Maximum deceleration, also known as negative acceleration, is the highest rate at which an object can slow down. It is measured in units of acceleration, such as meters per second squared (m/s^2).

4. How does maximum deceleration affect motion?

Maximum deceleration slows down the motion of an object, causing it to decrease in speed. This can also result in a change in direction if the object is moving in a curved path.

5. What are some real-life examples of constant acceleration and maximum deceleration?

Examples of constant acceleration include a car accelerating on a straight road, a rollercoaster moving down a slope, and a rocket launching into space. Examples of maximum deceleration include a car braking suddenly, a plane landing on a runway, and a skydiver slowing down as they approach the ground.

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