Two cars are traveling along a straight-line in the same direction, the lead car at 25.0 m/s and the other car at 32.0 m/s. At the moment the cars are 40.0 m apart, the lead driver applies the brakes, causing his car to have an acceleration of -2.20 m/s2.
(a) How much time does it take for the lead car to stop?
s
(b) Assuming that the chasing car brakes at the same time as the lead car, what must be the chasing car's minimum negative acceleration so as not to hit the lead car?
m/s2
(c) How much time does it take for the chasing car to stop?
s
See 9-22-11,5:59am post.
a)12.5 s
b)-13.6
c) 2.57 s
To solve this problem, we can use the equations of motion to determine the time required for each car to stop.
(a) To find the time it takes for the lead car to stop, we can use the equation:
v = u + at
where v is the final velocity (0 m/s in this case), u is the initial velocity (25.0 m/s in this case), a is the acceleration (-2.20 m/s²), and t is the time.
Rearranging the equation, we have:
t = (v - u) / a
Substituting the given values, we get:
t = (0 m/s - 25.0 m/s) / -2.20 m/s²
t = -25.0 m/s / -2.20 m/s²
t ≈ 11.36 s
Therefore, it takes approximately 11.36 seconds for the lead car to stop.
(b) For the chasing car to avoid hitting the lead car, its deceleration must be such that it is able to stop before reaching the lead car.
Since the chasing car is decelerating, its acceleration will also be negative. Let's denote the acceleration of the chasing car as -a' (negative of a prime).
To ensure that the chasing car does not hit the lead car, the distance traveled by the chasing car during braking should be equal to or greater than the initial separation between the two cars (40.0 m in this case).
We can use the equation:
s = ut + (1/2)at²
where s is the displacement (40.0 m in this case), u is the initial velocity (32.0 m/s in this case), t is the time, and a is the acceleration (-2.20 m/s²).
Substituting the values, we get:
40.0 m = (32.0 m/s)(t) + (1/2)(-a')(t)²
Simplifying further,
40.0 m = 32.0 m/s × t - (1/2) a' t²
Rearranging the terms,
(1/2) a' t² - 32.0 m/s × t + 40.0 m = 0
This is a quadratic equation in terms of t. Solving for t using the quadratic formula gives:
t = [32.0 m/s ± √(32.0 m/s)² - 4 × (1/2) a' × 40.0 m] / (1/2) a'
Simplifying,
t = 64.0 m/s ± √(1024.0 m²/s² + 80.0 m a')
Since the time cannot be negative, we take the positive value:
t = 64.0 m/s + √(1024.0 m²/s² + 80.0 m a')
For the chasing car to stop without hitting the lead car, the time taken by the chasing car (t) must be less than or equal to the time taken by the lead car (11.36 s).
Therefore, we can set up the inequality:
64.0 m/s + √(1024.0 m²/s² + 80.0 m a') ≤ 11.36 s
Solving this inequality will give us the range of possible values for the chasing car's acceleration (-a').
(c) To determine how much time it takes for the chasing car to stop, we can substitute the obtained value of -a' into the equation:
t = 64.0 m/s + √(1024.0 m²/s² + 80.0 m a')
This will give us the time required for the chasing car to stop.