A car traveling at 20m/s when the driver sees a child standing in the road. He takes 0.80s to react, then steps on the brakes and slows at 7.0m/s^2. How far does the car go before it stops?
d = Vo*t + (V^2-Vo^2)/2a
d = 20*0.8 + (0-20^2)/-14 = 44.6 m.
So is this finding the displacement or distance
Why is the value .8 seconds instead of 2.86
Well, the car certainly took its sweet time to make a decision there! Let's calculate how far it went before it finally stopped.
First, we need to calculate the distance the car traveled during the driver's reaction time. Since the car was traveling at a constant speed of 20 m/s for 0.80 seconds, the distance it traveled during that time can be calculated with the formula: distance = speed × time.
So, the distance traveled during reaction time is 20 m/s × 0.80 s = 16 meters.
Now, let's calculate the distance the car traveled while decelerating until it eventually stops. To do that, we'll use the formula: distance = initial velocity × time + (1/2) × acceleration × time squared.
In this case, the initial velocity is 20 m/s (as the car was traveling at that speed before braking), the acceleration is -7.0 m/s^2 (negative because the car is decelerating), and we need to find the time it takes for the car to stop. Since the car stops, its final velocity is 0 m/s.
Using the formula v^2 = u^2 + 2as, where v is final velocity, u is initial velocity, a is acceleration, and s is distance, we can solve for time.
Since v^2 = 0^2 and u = 20 m/s, we can rearrange the equation as 0 = 20^2 + 2(-7.0)s.
Simplifying this equation, we get -20^2 = 2(-7.0)s.
400 = -14s.
Finally, solving for s, we get s = -400/-14 ≈ 28.6 meters.
Therefore, the car travels approximately 28.6 meters during deceleration until it comes to a stop.
Adding the distance covered during the driver's reaction time, the total distance the car travels before stopping is approximately 16 meters + 28.6 meters = 44.6 meters.
That's quite a lengthy journey! Hopefully, the child managed to get out of the way on time!
To solve this problem, we can use the equations of motion. The important equations are:
1. v = u + at
2. s = ut + (1/2)at^2
3. v^2 = u^2 + 2as
Where:
- v is the final velocity
- u is the initial velocity
- a is the acceleration
- t is the time taken
- s is the distance traveled
Let's solve this step by step:
1. First, let's calculate the time it takes for the car to stop after the driver reacts. We are given that the driver takes 0.80 seconds to react, so t = 0.80s.
2. Next, let's find the final velocity (v) at the point the car stops. We know that the car slows down at an acceleration of -7.0 m/s^2 (negative because it's decelerating). The initial velocity (u) is 20 m/s. Using equation (1), we can plug in the values and solve for v:
v = u + at
v = 20 m/s + (-7.0 m/s^2)(0.80 s)
v = 20 m/s - 5.6 m/s
v = 14.4 m/s
3. Now, we can calculate the distance traveled (s) before the car comes to a stop. Using equation (3), we can plug in the values and solve for s:
v^2 = u^2 + 2as
(14.4 m/s)^2 = (20 m/s)^2 + 2(-7.0 m/s^2)s
207.36 m^2/s^2 = 400 m^2/s^2 + (-14.0 m/s^2)s
0 = 192.64 m^2/s^2 + (-14.0 m/s^2)s
Rearranging the equation:
(-14.0 m/s^2)s = -192.64 m^2/s^2
s = (-192.64 m^2/s^2) / (-14.0 m/s^2)
s ≈ 13.76 m
Therefore, the car will go approximately 13.76 meters before it comes to a stop.