To stop a car, you require first a certain reaction time to begin braking. Then the car slows under the constant braking deceleration. Suppose that the total distance moved by your car during these two phases is 56.7 m when its initial speed is 80.0 km/h, and 24.4 m when the initial speed is 47.8 km/h.
(a) What is your reaction time?
(b) What is the magnitude of the deceleration?
I tried eliminating the values in order to find accel and time, but the values canceled out and when I tried separating them, I got the answer wrong. What am I doing wrong?
I don't see how it is solvable, unless you assume the acceleration is different on each, or the time during deacceleartion is the same.
You have as variables: a, time to slow, tr.
Forgot to put my question at the end of it XD
This is from me trying to interpret what you've done in the previous times this question has been asked, but obviously, I'm wrong somewhere. At what point in the work am I messing up?
2.53 seconds
To solve this problem, you can use the equations of motion under constant acceleration. Let's break down the steps to find the answers.
First, we need to convert the initial speeds from km/h to m/s:
Initial speed 1: 80.0 km/h = (80.0 * 1000) / 3600 m/s ≈ 22.22 m/s
Initial speed 2: 47.8 km/h = (47.8 * 1000) / 3600 m/s ≈ 13.28 m/s
(a) To find the reaction time, we use the equation:
Distance = Initial Speed * Time + (1/2) * Acceleration * Time^2
For the first phase, when the car covers a distance of 56.7 m at an initial speed of 22.22 m/s, we can plug in the values:
56.7 = 22.22t + (1/2)a * t^2
For the second phase, when the car covers a distance of 24.4 m at an initial speed of 13.28 m/s, we again use the same equation:
24.4 = 13.28t + (1/2)a * t^2
Now we have two equations with two unknowns, namely the reaction time (t) and the acceleration (a).
To eliminate the time (t) variable, you can subtract the second equation from the first equation:
56.7 - 24.4 = 22.22t - 13.28t + (1/2)a * t^2 - (1/2)a * t^2
32.3 = 8.94t
Now solve for t:
t = 32.3 / 8.94 ≈ 3.61 seconds
So, the reaction time is approximately 3.61 seconds.
(b) To find the magnitude of the deceleration (a), we can substitute the value of t back into either of the equations we derived initially.
Let's use the first equation:
56.7 = 22.22 * 3.61 + (1/2)a * (3.61)^2
Now solve for a:
a = (56.7 - 22.22 * 3.61) / (1/2 * (3.61)^2) ≈ -3.91 m/s^2
The negative sign indicates deceleration.
So, the magnitude of the deceleration is approximately 3.91 m/s^2.
It's crucial to ensure you correctly input the values into the equations, with attention to unit conversions and signs, to obtain the accurate results.
Let me try to clarify
v1=22.222 m/s
v2=13.278 m/s
x=vi(t)+1/2a(t^2)
56.7=22.222t+1/2a(t^2)
24.4=13.278t+1/2a(t^2).
Through elimination
32.3=8.94t
t=3.611s
From there I separated to -vi=at as vf is 0
-22.222=a(3.611)
-13.278=a(3.611)
-35.5=7.222a
a=-4.92 m/s^2