Two cars are traveling along a straight line in the same direction, the lead car at 24.5m/s and the other car at 29.8m/s. At the moment the cars are 41.4m apart, the lead driver applies the brakes, causing her car to have an acceleration of -2.05m/s2.
A) How long does it take for the lead car to stop?
B) What is the distance it travels during this time?
C)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?
DHow long does it take for the chasing car to stop?
A. t = (Vf-Vo)/a,
t = (0-24.5) / -2.05 = 11.95 s.
B. d = (Vf^2-Vo^2)/2a,
d = (0-(24.5)^2) / -4.1 = 146.4 m.
C. a = (Vf^2-Vo^2)/2d,
a = (0-(29.8)^2) / 82.8 = -10.7 m/s^2.
D. t = (Vf-Vo)/a,
t = (0-29.8) / -10.7 = 2.78 s.
To solve these problems, we'll use the equations of motion. The four equations of motion are:
1) v = u + at
2) s = ut + 0.5at^2
3) v^2 = u^2 + 2as
4) s = ((u + v) / 2) * t
Where:
- v represents the final velocity
- u represents the initial velocity
- a represents the acceleration
- t represents time
- s represents the distance traveled
A) How long does it take for the lead car to stop?
In this case, the initial velocity of the lead car (u) is 24.5 m/s. The acceleration (a) is -2.05 m/s^2 because it's braking. The final velocity (v) will be 0 m/s because it stops. We need to find the time (t).
Using equation 1: v = u + at
0 = 24.5 + (-2.05)t
-24.5 = -2.05t
t = -24.5 / -2.05 ≈ 11.95 seconds
So, it takes approximately 11.95 seconds for the lead car to stop.
B) What is the distance it travels during this time?
To find the distance traveled, we still use equation 1 but solve for distance (s).
Using equation 1: v = u + at
0 = 24.5 + (-2.05)t
-24.5 = -2.05t
t = -24.5 / -2.05 ≈ 11.95 seconds
Now, using equation 2: s = ut + 0.5at^2
s = 24.5 * 11.95 + 0.5 * (-2.05) * (11.95)^2
s ≈ 145.84 meters
So, the lead car travels approximately 145.84 meters during the time it takes to stop.
C) 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?
To avoid hitting the lead car, the chasing car needs to decelerate at a rate that allows it to stop within the same distance as the lead car. Since the chasing car starts 41.4 meters behind the lead car, it needs to stop within this distance.
Using equation 3: v^2 = u^2 + 2as
0^2 = 29.8^2 + 2a(41.4)
0 = 888.04 + 82.8a
-82.8a = 888.04
a = 888.04 / -82.8 ≈ -10.73 m/s^2
So, the chasing car's minimum negative acceleration should be approximately -10.73 m/s^2.
D) How long does it take for the chasing car to stop?
To find the time it takes for the chasing car to stop, we'll use the same equation as in part A but with the new acceleration value.
Using equation 1: v = u + at
0 = 29.8 + (-10.73)t
-29.8 = -10.73t
t = -29.8 / -10.73 ≈ 2.77 seconds
So, it takes approximately 2.77 seconds for the chasing car to stop.