A car starting from rest& moves for 20 seconds its velocity reach 5m/s & moves at this velocity for 30 seconds.it's velocity increases to 10m/s over 10 seconds.it moves at this velocity for 40 seconds &then slows down to come to rest after further 60 seconds.how far did the car travel during this time?and during which period was the accleration of the car the lowest and during which period period was the accleration of the car the highest?
see the problem below this one.
for each stage, the distance s = v0*t + 1/2 at^2
I don't understand
Phase 1
v = Vi + a t
5 = 0 + a * 20
a = 1/4
x = 0 + 0 t + (1/2) a t^2
x = (1/8)(400) = 50 meters
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Phase 2
Xi = 50
Vi = 5 for next 30 s
a = 0
v = Vi + a t = Vi = 5
x = Xi + Vi t + (1/2) a t^2
x = 50 + 5 t = 50 + 5(30) = 200
Phase 3
Xi = 200
Vi = 5
v after 10 s = 10 m/s
v = Vi + a t
10 = 5 + a * 10
5 = 10 a
a = 0.5 m/s^2
x = Xi + Vi t +(1/2)a t^2
x = 200 + 5(10) + (1/2)(0.5)(100)
x = 200 + 50 + 25 = 275
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Phase 4
Xi = 275
Vi = 10 m/s
a = 0 for 40 seconds
x = Xi + 10*40 = 275 + 400 = 675 meters
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Phase 5
Xi = 275
Vi = 10 m/s
stops in 60 seconds
v =10 + a (60)
0 = 10 + 60 a
a = - (1/6) meter/second^2
x = 675 + 10(60) -(1/2)(1/6)(3600)
x = 675 + 600 - 300 = 975 meters for the whole trip
To calculate the total distance traveled by the car, we need to find the distance covered in each time period and then add them together.
1) First, let's calculate the distance during the initial period of acceleration. The car starts from rest and reaches a velocity of 5 m/s in 20 seconds. To find the distance traveled, we can use the formula:
Distance = (Initial velocity * Time) + (0.5 * Acceleration * Time^2)
Here, the initial velocity is 0 m/s, the time is 20 seconds, and the acceleration is unknown. As we need to calculate the acceleration, we rearrange the formula:
Acceleration = (Final velocity - Initial velocity) / Time
Acceleration = (5 m/s - 0 m/s) / 20 seconds = 0.25 m/s^2
Now we can calculate the distance covered during the initial acceleration:
Distance = (0 m/s * 20s) + (0.5 * 0.25 m/s^2 * (20s)^2) = 100 meters
So, the car traveled 100 meters during the initial period.
2) Next, let's calculate the distance traveled at a constant speed of 5 m/s for 30 seconds. Since the car is moving at a constant speed, there is no acceleration. Therefore, the distance is simply:
Distance = Velocity * Time = 5 m/s * 30 s = 150 meters
The car traveled 150 meters during this period.
3) Now let's calculate the distance covered during the second acceleration phase. The car starts with a velocity of 5 m/s and reaches a final velocity of 10 m/s in 10 seconds. We can use the same formula as before, but this time the acceleration is unknown:
Acceleration = (Final velocity - Initial velocity) / Time
Acceleration = (10 m/s - 5 m/s) / 10 seconds = 0.5 m/s^2
Now we can calculate the distance covered during the second acceleration:
Distance = (5 m/s * 10s) + (0.5 * 0.5 m/s^2 * (10s)^2) = 75 meters
The car traveled 75 meters during this period.
4) Finally, the car slows down and comes to rest over 60 seconds. Since it comes to rest, the final velocity is 0 m/s. Therefore, the distance covered during this period can be calculated using the same formula as before:
Distance = (Initial velocity * Time) + (0.5 * Acceleration * Time^2)
Here, the initial velocity is 10 m/s (from the previous period), the time is 60 seconds, and the acceleration is unknown. Rearranging the formula to solve for acceleration:
Acceleration = (Final velocity - Initial velocity) / Time
Acceleration = (0 m/s - 10 m/s) / 60 seconds = -0.166 m/s^2 (negative sign indicates deceleration)
Now we can calculate the distance covered during the deceleration:
Distance = (10 m/s * 60s) + (0.5 * -0.166 m/s^2 * (60s)^2) = 300 meters
The car traveled 300 meters during this period.
To find the total distance traveled, we sum up the distances from each period:
Total Distance = 100 m + 150 m + 75 m + 300 m = 625 meters
So, the car traveled a total of 625 meters.
Now, let's determine the periods of lowest and highest acceleration:
- The lowest acceleration is during the period of constant velocity, where the car travels at 5 m/s for 30 seconds. Since there is no change in velocity during this period, the acceleration is 0 m/s^2.
- The highest acceleration is during the second period of acceleration, where the velocity increases from 5 m/s to 10 m/s in 10 seconds. The previously calculated acceleration for this period was 0.5 m/s^2.
Therefore, the period of lowest acceleration is when the car moves at a constant velocity of 5 m/s for 30 seconds, and the period of highest acceleration is during the second acceleration phase when the velocity increases from 5 m/s to 10 m/s in 10 seconds.