Assume that the brakes in your car create a constant deceleration of 3.1 m/s2 regardless of how fast you are driving.
part A= If you double your driving speed from 17 m/s to 34 m/s , does the distance required to come to a stop increase by a factor of 2 or a factor of 4?
part B= Calculate the stopping distance for initial speed of 17 m/s .
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Part A: To determine whether the distance required to come to a stop doubles or quadruples when the driving speed doubles, we can use the kinematic equation for uniformly decelerated motion:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s when coming to a stop)
u = initial velocity (17 m/s)
a = acceleration (deceleration in this case, -3.1 m/s^2)
s = distance
We can rewrite the equation as:
0^2 = (2*17^2) + 2*(-3.1)*s
Simplifying this equation, we get:
0 = 578 - 6.2s
Rearranging the equation, we find:
6.2s = 578
s = 578 / 6.2
s ≈ 93.23 m
Now, let's calculate the stopping distance for a driving speed of 34 m/s.
0^2 = (2*34^2) + 2*(-3.1)*s'
0 = 2312 - 6.2s'
6.2s' = 2312
s' = 2312 / 6.2
s' ≈ 373.55 m
The difference between the stopping distances is:
s' - s = 373.55 - 93.23
≈ 280.32 m
Since the stopping distance increases by approximately 280.32 m when the driving speed doubles, the distance required to come to a stop increases by a factor of 280.32 / 93.23 ≈ 3.01.
Therefore, the distance required to come to a stop increases by a factor of approximately 3 (not 2).
Part B: Since we already have the calculation for the stopping distance when the initial speed is 17 m/s, the stopping distance is approximately 93.23 meters.
To determine the answers to both parts of the question, we can use the equations of linear motion.
Part A:
Let's assume the initial speed is v1 = 17 m/s and the final speed is v2 = 34 m/s. We are trying to find out if the stopping distance increases by a factor of 2 or a factor of 4.
First, let's find the time it takes to come to a stop. We can use the formula:
v2 = v1 + at
Plugging in the values:
34 m/s = 17 m/s + 3.1 m/s² * t
Solving for t:
t = (34 m/s - 17 m/s) / 3.1 m/s²
t = 17 m/s / 3.1 m/s²
t ≈ 5.48 s
Now, we can find the stopping distance using the formula:
d = v1 * t + 0.5 * a * t²
Plugging in the values:
d = 17 m/s * 5.48 s + 0.5 * 3.1 m/s² * (5.48 s)²
d ≈ 93.16 m
Now, let's double the initial speed to v1' = 2 * v1 = 2 * 17 m/s = 34 m/s and find the stopping distance, d'.
Using the same formulas:
t' = v2 / a = 34 m/s / 3.1 m/s² ≈ 10.97 s
d' = v1' * t' + 0.5 * a * t'²
d' = 34 m/s * 10.97 s + 0.5 * 3.1 m/s² * (10.97 s)²
d' ≈ 378.78 m
Therefore, the stopping distance increases by a factor of 378.78 m / 93.16 m ≈ 4.
Part B:
To calculate the stopping distance for an initial speed of 17 m/s, we can use the same formulas as before.
t = v2 / a = 17 m/s / 3.1 m/s² ≈ 5.48 s
d = v1 * t + 0.5 * a * t²
d = 17 m/s * 5.48 s + 0.5 * 3.1 m/s² * (5.48 s)²
d ≈ 93.16 m
Therefore, the stopping distance for an initial speed of 17 m/s is approximately 93.16 meters.
A. (2Vo)^2/Vo^2 = 4Vo^2/Vo^2 = 4.
B. V ^2 = Vo^2+2a*d = 0
17^2-6.2d = 0.
d = 46.6 m.