You are driving to the grocery store at 14.8 m/s. You are 140.0 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.440 s and that your car brakes with constant acceleration. You are 133 m from the intersection when you begin to apply the brakes.
a) What acceleration will bring you to rest as you just reach the intersection?
b) How long does it take you to stop?
Ok, during the reaction time, he moves
14.8*.440 m
subtract that from 133 to get stopping distance.
Vf^2=Vi^2+2ad
solve for a
Time? Vf=Vi+at
solve for t.
i do not think that is correct because i am getting the incorrect answer.
Kinematics Equations:
Vx1 = Vx0 + axΔt
(Vx1)^2 = (Vx0)^2 + 2axΔx
Δx = Vx0Δt + ½ ax(Δt^2)
Had the distance from the intersection when braking begins not been given:
Remaining distance = 140.0 m
Current Velocity = 14.8 m/s
Reaction time = 0.440 s
Find distance from intersection when breaking begins:
140.0 m - [14.8 m/s (0.440 s)] = 133.488
= 133. 488 m.
Find what acceleration will bring you to rest as you just reach the intersection:
Use the second kinematics equation, rearrange the variables to get "ax" by itself on one side:
(Vx1)^2 = (Vx0)^2 + 2axΔx
= V(Vx1)^2 - (Vx0)^2 = 2axΔx
= [1/(2Δx)] [(Vx1)^2 - (Vx0)^2] = ax
So: ax = [1/(2Δx)] [(Vx1)^2 - (Vx0)^2]
Now plug in quantities for the known variables:
ax = Brake acceleration = ? m/s^2
Δx = Distance to stop = 133.488 m
Vx1 = Final Velocity = 0.00 m/s
Vx0 = Initial Velocity = 14.8 m/s
ax = {1/[2(133.488 m)]} {[0.00 m/s]^2 -[14.8 m/s]^2}
ax = -0.82044828 m/s^2
Final answer (significant figures):
ax = -0.820 m/s^2
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Use the first kinematics equation to determine how long it will take you to stop:
Vx1 = Vx0 + axΔt
You can rearrange the equation to get Δt by itself, or just plug in what you know:
0.00 m/s = 14.8 m/s + [(-0.82044828 m/s^2)(Δt)
-14.8 m/s = (-0.82044828 m/s^2)(Δt)
Δt = (-14.8 m/s)/(-0.82044828)
Δt = 12.14263454
Final answer (significant figures):
Δt = 12.1 s
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Hope that helped- keep those kinematic equations handy (or, memorize them)!
To solve this problem, we can use the following kinematic equations:
1) Distance traveled during acceleration:
d = v0t + (1/2)at^2
2) Final velocity when coming to a stop:
v = v0 + at
We are given the following information:
- Initial velocity (v0) = 14.8 m/s
- Initial distance from the intersection (d0) = 140.0 m
- Distance traveled while applying the brakes (d1) = 133.0 m
- Reaction time (t_r) = 0.440 s
a) To calculate the acceleration (a) that will bring you to rest as you just reach the intersection:
We need to find the time it takes to reach the intersection when the brakes are applied.
1) Calculate the time taken to reach the intersection while driving at a constant speed:
d0 = v0t1 => 140.0 m = 14.8 m/s * t1 => t1 = 9.46 seconds
2) Subtract the reaction time from the total time:
t = t1 - t_r => t = 9.46 s - 0.440 s => t = 9.02 seconds
3) Calculate the deceleration (negative acceleration) during braking using the formula:
a = (2 * d1 - v0 * t) / t^2
= (2 * 133.0 m - 14.8 m/s * 9.02 s) / (9.02 s)^2
b) To calculate the time it takes to stop:
We need to find the time it takes for the velocity to become zero after the brakes are applied.
1) Calculate the time taken for the velocity to reach zero:
v = 0 = v0 + at_stop
t_stop = -v0 / a
2) Substitute the values into the equation:
t_stop = -14.8 m/s / (a from part a)
Now we can calculate the values.
a) Using the above formula, insert the values to find the acceleration:
a = (2 * 133.0 m - 14.8 m/s * 9.02 s) / (9.02 s)^2
b) Using the formula above, insert the values found in part a to find the time taken to stop:
t_stop = -14.8 m/s / (a from part a)
Solving these equations will give you the desired answers.