Two cars are traveling along a straight line in the same direction, the lead car at 26 m/s and the other car at 40 m/s. At the moment the cars are 49 m apart, the lead driver applies the brakes, causing the car to have a deceleration of 1.5 m/s^2. How long does it take for the lead car to stop? Answer in units of s. (52/3)
032 (part 2 of 3) 10.0 points
Assume that the driver of the chasing car applies the brakes at the same time as the driver of the lead car. What must the chasing car’s minimum negative acceleration be to avoid hitting the lead car? Answer in units of m/s^2.
How long does it take the chasing car to
stop? Answer in units of s
1. t = (Vf - Vo) / a,
t = (0 - 26) / -1.5 = 17.33s.
d = (Vf^2 - Vo^2) / 2a,
d = (0 - (26)^2) / -3 = 225.33m = stopping distance for lead car.
d = 49 + 225.33 = 274.33m = Required stopping distance for the chasing car.
a = (Vf^2 - Vo^2) / 2d,
a = (0 - (40)^2) / 548.66 = -2.92m/s^2.
= required deceleration of 2nd car.
t = (Vf - Vo) / a,
t = (0 - 40) / -2.92 = 13.72s. = Time it takes for the 2nd car to stop.
To find the time it takes for the lead car to stop, we can use the equation:
v = u + at
Where:
v = final velocity (0 m/s, since the car stops)
u = initial velocity of the lead car (26 m/s)
a = deceleration of the lead car (-1.5 m/s^2)
t = time
Rearranging the equation to solve for t, we have:
t = (v - u) / a
Plugging in the values, we get:
t = (0 - 26) / (-1.5)
t = 52 / 3
Therefore, it takes the lead car 52/3 seconds to stop.
Now let's move on to the second part of the question.
To calculate the minimum negative acceleration the chasing car must have to avoid hitting the lead car, we can use the equation:
s = ut + 0.5at^2
Where:
s = distance between the two cars (49 m)
u = initial velocity of the chasing car (40 m/s, as it doesn't brake initially)
a = acceleration of the chasing car (what we need to find)
t = time it takes for the lead car to stop (52/3 s)
Rearranging the equation and solving for a, we have:
a = 2s / t^2
Plugging in the values, we get:
a = 2 * 49 / (52/3)^2
a ≈ -6.42 m/s^2 (rounded to two decimal places)
Therefore, the chasing car's minimum negative acceleration must be approximately -6.42 m/s^2 to avoid hitting the lead car.
Finally, to find out how long it takes for the chasing car to stop, we use the same formula as before:
v = u + at
Where:
v = final velocity (0 m/s, since the car stops)
u = initial velocity of the chasing car (40 m/s)
a = deceleration of the chasing car (what we need to find)
t = time
Rearranging the equation to solve for t, we have:
t = (v - u) / a
Plugging in the values, we get:
t = (0 - 40) / a
Since we have found the minimum negative acceleration (approximately -6.42 m/s^2) in the previous step, we can now substitute that value in:
t = (0 - 40) / (-6.42)
t ≈ 6.23 s (rounded to two decimal places)
Therefore, it takes the chasing car approximately 6.23 seconds to stop.
To find the minimum negative acceleration of the chasing car to avoid hitting the lead car, we need to analyze the relative motion between the two cars.
First, let's find the time it takes for the lead car to stop. We know its initial velocity (v1) is 26 m/s, and it decelerates at a rate of 1.5 m/s^2 until it stops. We can use the equation:
v2 = v1 + at
where v2 is the final velocity, v1 is the initial velocity, a is the acceleration, and t is the time. In this case, v2 is 0 m/s (since the car stops). We can rearrange the equation to solve for t:
0 = 26 + (1.5)t
-26 = (1.5)t
t = -26 / 1.5
t ≈ -17.33 s
The time it takes for the lead car to stop is approximately 17.33 seconds.
Now let's analyze the motion of the chasing car. Since it starts 49 m behind the lead car and both cars decelerate at the same time, the chasing car needs to have enough negative acceleration to stop before hitting the lead car.
Let's set up an equation to solve for the minimum negative acceleration of the chasing car. We can use the equation:
x = v1t + (1/2)at^2
where x is the initial distance between the cars (49 m), v1 is the initial velocity of the chasing car (40 m/s), a is the acceleration of the chasing car, and t is the time it takes to stop. Since the chasing car starts from rest, its initial velocity is 0 m/s.
49 = 0 + (1/2)a(t^2)
49 = (1/2)a(t^2)
Dividing both sides by (1/2)t^2:
98 / t^2 = a
The minimum negative acceleration (a) of the chasing car is given by the equation 98 / t^2.
Now we need to find the time it takes for the chasing car to stop. Since both cars decelerate for the same time until the lead car stops, the time it takes for the chasing car to stop will also be approximately 17.33 seconds.
Plugging this value into the equation for acceleration:
a = 98 / (17.33)^2
a ≈ 0.337 m/s^2
Therefore, the chasing car's minimum negative acceleration to avoid hitting the lead car is approximately -0.337 m/s^2.
The time it takes for the chasing car to stop is also approximately 17.33 seconds.