A car's initial velocity of 13.5 m/s accelerates at a rate of 1.9 m/s^2 for 6.2 s. Then, it accelerates at a rate of -1.2 m/s^2 until it stops.
1) What is the car's maximum speed?
2) What is the total time from the start of the first acceleration until the car is stopped?
3) Find the total distance that the car traveled.
Please include all steps used in the calculations if possible. Thank you!
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To find the answers to the questions, we can use the equations of motion. Let's go step by step.
1) What is the car's maximum speed?
To find the maximum speed of the car, we need to find the velocity at the end of the first acceleration and then calculate the final velocity when it stops.
The initial velocity of the car, u, is 13.5 m/s.
The acceleration during the first period, a1, is 1.9 m/s².
The time taken during the first period, t1, is 6.2 seconds.
Using the equation of motion:
v = u + a * t
The final velocity after the first acceleration is:
v1 = u + a1 * t1
= 13.5 m/s + 1.9 m/s² * 6.2 s
= 13.5 m/s + 11.78 m/s
= 25.28 m/s
Now, during the second period, the acceleration is -1.2 m/s² until the car stops. Let's denote the final velocity as v2.
Using the equation of motion:
v = u + a * t
Since the car stops, the final velocity is 0 m/s.
v2 = u + a2 * t2
0 = 25.28 m/s + (-1.2 m/s²) * t2
Solving for t2:
1.2 m/s² * t2 = 25.28 m/s
t2 = 25.28 m/s / 1.2 m/s²
t2 = 21.07 seconds
Now, let's substitute this value back into the equation for v2 to find the maximum speed:
v2 = u + a2 * t2
= 13.5 m/s + (-1.2 m/s²) * 21.07 s
= 13.5 m/s - 25.284 m/s
= - 11.784 m/s (negative sign indicates the car is traveling in the opposite direction)
Therefore, the maximum speed of the car is 11.784 m/s.
2) What is the total time from the start of the first acceleration until the car is stopped?
The total time is the summation of the time it takes during the first acceleration and the time it takes during the second acceleration (until the car stops).
Total time = t1 + t2
= 6.2 seconds + 21.07 seconds
= 27.27 seconds
Therefore, the total time from the start of the first acceleration until the car is stopped is 27.27 seconds.
3) Find the total distance that the car traveled.
To find the total distance traveled by the car, we need to find the distance traveled during the first acceleration and the distance traveled during the second acceleration (until the car stops).
The distance traveled during uniform acceleration can be calculated using the equation:
s = ut + (1/2) a t^2
During the first acceleration:
s1 = (1/2) * a1 * t1^2
= (1/2) * 1.9 m/s² * (6.2 s)^2
= (1/2) * 1.9 m/s² * 38.44 s²
= 36.6348 m
During the second acceleration:
s2 = (1/2) * a2 * t2^2
= (1/2) * (-1.2 m/s²) * (21.07 s)^2
= (1/2) * (-1.2 m/s²) * 443.9449 s²
= - 265.1669 m (negative sign indicates the direction of travel)
The total distance traveled is the sum of s1 and s2:
Total distance = s1 + s2
= 36.6348 m - 265.1669 m
= -228.5321 m
The negative sign indicates the car traveled in the opposite direction.
Therefore, the total distance that the car traveled is 228.5321 m.
To answer these questions, we will need to use the equations of motion, specifically the equations for uniform acceleration. Let's break down each question and solve them step by step:
1) What is the car's maximum speed?
First, we need to find the speed of the car after the first acceleration. We can use the equation:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Given that the initial velocity (u) is 13.5 m/s, the acceleration (a) is 1.9 m/s^2, and the time (t) is 6.2 s, we can substitute these values into the equation:
v1 = 13.5 m/s + (1.9 m/s^2)(6.2 s)
v1 = 13.5 m/s + 11.78 m/s
v1 = 25.28 m/s
The car's speed after the first acceleration is 25.28 m/s.
Next, we need to find the speed of the car when it stops after the second acceleration. Since the acceleration is negative (-1.2 m/s^2), we can use the equation:
v^2 = u^2 + 2as
where:
v = final velocity,
u = initial velocity,
a = acceleration (negative value since deceleration),
s = distance
Given that the initial velocity (u) is 25.28 m/s and the acceleration (a) is -1.2 m/s^2, we can substitute these values into the equation:
0 = (25.28 m/s)^2 + 2(-1.2 m/s^2)s
0 = 638.51 m^2/s^2 - 2.4 m/s^2s
2.4 m/s^2s = 638.51 m^2/s^2
s = 638.51 m^2/s^2 / 2.4 m/s^2
s = 266.046 m
The distance covered during deceleration is 266.046 m.
To find the time taken for deceleration, we can use the equation:
v = u + at
Since we want to find the time (t), we can rearrange this equation:
t = (v - u) / a
Given that the initial velocity (u) is 25.28 m/s, the final velocity (v) is 0 m/s, and the acceleration (a) is -1.2 m/s^2, we can substitute these values into the equation:
t = (0 m/s - 25.28 m/s) / (-1.2 m/s^2)
t = (-25.28 m/s) / (-1.2 m/s^2)
t = 21.067 s
The time taken for deceleration is 21.067 s.
Since we know the time taken for the first acceleration is 6.2 s and the time taken for deceleration is 21.067 s, we can calculate the total time from the start of the first acceleration until the car stops:
Total time = 6.2 s + 21.067 s
Total time = 27.267 s
The total time from the start of the first acceleration until the car stops is 27.267 s.
3) Find the total distance that the car traveled.
To find the total distance traveled, we need to sum up the distances covered during the first acceleration and the deceleration:
Distance = Distance1 + Distance2
Distance = 0.5(u + v1)t1 + 0.5(v1 + v2)t2
Given that:
u = 13.5 m/s (initial velocity)
v1 = 25.28 m/s (velocity after the first acceleration)
t1 = 6.2 s (time for the first acceleration)
v2 = 0 m/s (final velocity after deceleration)
t2 = 21.067 s (time for deceleration)
Distance = 0.5(13.5 m/s + 25.28 m/s)(6.2 s) + 0.5(25.28 m/s + 0 m/s)(21.067 s)
Distance = 0.5(38.78 m/s)(6.2 s) + 0.5(25.28 m/s)(21.067 s)
Distance = 120.458 m + 266.046 m
Distance = 386.504 m
The total distance that the car traveled is 386.504 m.
To summarize:
1) The car's maximum speed is 25.28 m/s.
2) The total time from the start of the first acceleration until the car stops is 27.267 s.
3) The total distance that the car traveled is 386.504 m.