A car traveling 51.1 km/h is 29.6 m from a barrier when the driver slams on the brakes. The car hits the barrier 3.08 s later. (a) What is the car's constant deceleration magnitude before impact? (b) How fast is the car traveling at impact?
See Related Questions: Wed, 9-2 09, 11:44 PM.
To solve this problem, we can use the equations of kinematics and the principles of constant acceleration. Here are the steps to find the answers to (a) and (b):
(a) To find the car's constant deceleration magnitude, we can use the equation of motion:
vf = vi + at
where vf is the final velocity, vi is the initial velocity, a is the acceleration, and t is the time taken.
Given:
Initial velocity (vi) = 51.1 km/h
Initial velocity in m/s = 51.1 km/h * (1000 m/1 km) * (1 h/3600 s) = 14.2 m/s
Time (t) = 3.08 s
Final velocity (vf) = 0 m/s (since the car comes to rest)
Substituting the known values into the equation above, we can solve for acceleration (a):
0 = 14.2 m/s + a * 3.08 s
Rearranging the equation to isolate the acceleration (a), we get:
a = -14.2 m/s ÷ 3.08 s
Evaluating this expression, we find that the car's constant deceleration magnitude before impact is approximately -4.61 m/s² (negative since it is deceleration).
(b) To find the speed of the car at impact, we can use the equation of motion:
vf^2 = vi^2 + 2aΔx
where Δx is the distance.
Given:
Δx = 29.6 m
vi = 14.2 m/s
a = -4.61 m/s²
Substituting the known values into the equation, we can solve for vf:
vf^2 = (14.2 m/s)^2 + 2(-4.61 m/s²)(29.6 m)
Simplifying this equation, we get:
vf^2 = 201.64 m²/s² + (-271.9 m²/s²)
vf^2 = -70.26 m²/s²
Finally, we take the square root of both sides to solve for vf:
vf = √(-70.26 m²/s²)
Since the velocity cannot be imaginary in this case, we can conclude that the car does not hit the barrier (as the negative sign suggests) and therefore, the answer is that the car did not hit the barrier.
Thus, the car is traveling at a speed of approximately 14.2 m/s before impact, and it does not hit the barrier.