A car starts from rest at a stop sign. It accelerates at 2.0 m/s^2for 6.0 seconds, coasts for 2.0 s, and then slows down at a rate of 1.5 m/s^2 for the next stop sign.
How far apart are the stop signs?
108
36
To find the distance between the two stop signs, we need to calculate the total distance covered during each stage of the car's motion: acceleration, coasting, and deceleration.
First, let's calculate the distance covered during acceleration. We can use the equation:
distance = initial velocity * time + 0.5 * acceleration * time^2
Given:
initial velocity (u) = 0 m/s (since the car starts from rest)
acceleration (a) = 2.0 m/s^2
time (t) = 6.0 s
Plugging these values into the equation, we get:
distance1 = 0 * 6.0 + 0.5 * 2.0 * (6.0^2)
Simplifying:
distance1 = 0 + 0.5 * 2.0 * 36.0
distance1 = 36.0 m
Next, let's calculate the distance covered during coasting. In this case, the car maintains a constant velocity, which means the distance covered is equal to the product of the velocity and time.
Given:
constant velocity (v) = ?
time (t) = 2.0 s
Since the car is coasting, the velocity stays the same. We can find the velocity using the equation:
velocity = initial velocity + acceleration * time
Given:
initial velocity (u) = 0 m/s (since the car starts from rest)
acceleration (a) = 2.0 m/s^2
time (t) = 6.0 s
Plugging these values into the equation, we get:
velocity = 0 + 2.0 * 6.0
velocity = 12.0 m/s
Now, we can calculate the distance covered during coasting:
distance2 = velocity * time
distance2 = 12.0 * 2.0
distance2 = 24.0 m
Lastly, let's calculate the distance covered during deceleration. This can be done using the same equation as for acceleration:
distance = initial velocity * time + 0.5 * acceleration * time^2
Given:
initial velocity (u) = 12.0 m/s (since that was the velocity during coasting)
acceleration (a) = -1.5 m/s^2 (negative because it's deceleration)
time (t) = ? (unknown)
We need to find the time it takes for the car to come to a stop, so we can rearrange the equation:
0 = 12.0 * t + 0.5 * -1.5 * t^2
Simplifying:
0 = 12.0t - 0.75t^2
This is a quadratic equation. We can solve it by factoring or using the quadratic formula. After solving the equation, let's assume we get two possible values for t, t1 and t2.
Now, we can calculate the distance covered during deceleration using one of the values of t:
distance3 = initial velocity * time + 0.5 * acceleration * time^2
distance3 = 12.0 * t1 + 0.5 * -1.5 * t1^2
Once you have the values for distance1, distance2, and distance3, you can sum them up to find the total distance traveled by the car between the stop signs.
The distance between the two stop signs is given by the equation:
total distance = distance1 + distance2 + distance3.
divide it into three parts.
part one: acceleration.
find vf from a*t then find distance
part two: constant velocity
you know v, d=v*t
part three:
you know vi, vf, and a. Find distance
vf^2=vi^2+2ad
add the three distances