A car moves along an x axis through a distance of 900 m, starting at rest (at x = 0) and ending at
rest (at x = 900 m). Through the first ¼ of that distance, its acceleration is +2.25 m/s2
. Through
the rest of that distance, its acceleration is – 0.750 m/s2
. Find its maximum speed?
You have to tackle these type of problem in steps: Solve for the first leg of the trip, and then solve for the second leg of the trip.
Use the following equation for the first leg of the trip:
Vf^2=Vi^2+2ad
Where
d=900m*(1/4)=225m
a=2.2 m/s^2
Vi=0m/s
and
Vf=?
Solve for Vf:
Vf^2=Vi^2+2ad
Vf^2=0+2(2.25m/s^2)*(225m)
Vf^2=1,012.5m^2/s^2
Vf=(1,012.5m^2/s^2)^1/2
Vf=31.8m/s^2
Vf for the first leg of the trip =Vi for the second leg of the trip. Solve for Vf of the second leg of the trip, which is the maximum speed of the trip.
Again, use the following equation for the first leg of the trip:
Vf^2=Vi^2+2ad
Where
d=900m*(3/4)=675m
a=0.750 m/s2
Vi=31.8m/s^2
and
Vf=?
Solve for Vf:
Vf^2=Vi^2+2ad
Vf^2=31.8m/s^2+2(0.750 m/s2)*(675m)
I'll let you do the rest.
To find the maximum speed of the car, we can divide the problem into two parts:
1. First ¼ of the distance:
In this part, the car has a constant acceleration of +2.25 m/s^2. We can use the equation:
v^2 = u^2 + 2as,
where v is the final velocity, u is the initial velocity (0 m/s since the car starts at rest), a is the acceleration, and s is the distance.
Plugging in the values:
v^2 = 0 + 2 * 2.25 * (1/4 * 900)
v^2 = 2.25 * 225
v^2 = 506.25
v = √506.25
v ≈ 22.5 m/s
2. Remaining distance:
In this part, the car has a constant acceleration of -0.750 m/s^2. Again, using the same equation:
v^2 = u^2 + 2as,
where v is the final velocity, u is the initial velocity (22.5 m/s since the car reached that speed in the previous part), a is the acceleration, and s is the remaining distance (3/4 of 900 m).
Plugging in the values:
v^2 = 22.5^2 + 2 * -0.750 * (3/4 * 900)
v^2 = 506.25 - 1012.50
v^2 = -506.25
Since velocity cannot be negative, we can ignore this part and take the maximum speed as 22.5 m/s.
Therefore, the maximum speed of the car is 22.5 m/s.
To find the maximum speed of the car, we need to consider two parts of its motion: the first quarter of the distance with an acceleration of +2.25 m/s^2, and the rest of the distance with an acceleration of -0.750 m/s^2.
First, let's find the time it takes for the car to reach the end of the first quarter (225 m) using the first quarter's acceleration:
1. We can use the equation of motion: v = u + at, where
- v is the final velocity (unknown)
- u is the initial velocity (0 m/s, because the car starts at rest)
- a is the acceleration (+2.25 m/s^2)
- t is the time
Therefore, the equation becomes v = 0 + (2.25 m/s^2) * t.
2. We can also use the equation of motion: s = ut + 0.5at^2, where
- s is the distance (225 m)
- u is the initial velocity (0 m/s)
- a is the acceleration (+2.25 m/s^2)
- t is the time
Substituting the values, the equation becomes 225 m = 0.5 * (2.25 m/s^2) * t^2.
3. We have two equations involving 't', so we can solve them simultaneously:
v = 2.25t
225 = 1.125t^2
Rearranging the second equation to isolate 't', we have t^2 = 225 / 1.125. Solving that, we get t ≈ 15.87 seconds.
Next, let's find the maximum speed when the car decelerates with an acceleration of -0.750 m/s^2:
1. Using the equation v = u + at, and taking u to be the velocity at the end of the first quarter (2.25t) and 'v' to be the maximum speed, we can write:
v = 2.25t - (0.750 m/s^2) * t
2. We set the total distance (900 m) equal to the sum of the distances for each quarter:
900 m = 4 * (225 m)
3. At the end of the first quarter, the velocity is 2.25t, and at the end of the fourth quarter (900 m), the velocity is 0 m/s.
4. We can use the equation of motion: v^2 = u^2 + 2as, where
- v is the final velocity (0 m/s)
- u is the initial velocity (2.25t)
- a is the acceleration (-0.750 m/s^2)
- s is the distance (900 m - 225 m = 675 m)
Writing this equation, we get:
0 = (2.25t)^2 + 2 * (-0.750 m/s^2) * 675 m
5. Solving the equation, we find that (2.25t)^2 ≈ 1012.50 m^2/s^2.
Taking the square root of both sides:
2.25t ≈ √(1012.50 m^2/s^2)
Simplifying further, we get 2.25t ≈ 31.87 m/s.
Finally, we can substitute the value of t back into the equation (2.25t) to find the maximum speed of the car. Approximating it, we get:
v ≈ 2.25 * 15.87 ≈ 35.82 m/s.
Therefore, the maximum speed of the car is approximately 35.82 m/s.