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A car start moving along a line first with acceleration a=2 starting from rest then uniformly and finally decelerating at the same rate and comes to rest the total time of motion is 10s the average speed during the time is 3.2m/s how long does the car moved uniformly?
Verified Answer
A car start moving along a line first with acceleration a=2 starting f...
Let the total distance travelled by the car be D. The total time of motion is 10s; therefore the average speed is (D/10) m/s. We are given the average speed as 3.2 m/s. So (D/10) = 3.2 or D= 32 m.
Now the car starts from rest, accelerates for some time at 2 m/s^2, then moves at a uniform speed and finally decelerates at 2m/s^2 to come to rest. As the car had started from rest, the time for acceleration and deceleration is the same. Let the time the car accelerates and decelerates be t1 and the time it moves at a uniform speed be t2. We get: 2*t1+ t2 = 10.
Now the distance it travels while accelerating is 2*t1*t1/2, this is the same as the distance it travels while decelerating. The speed it has after t1 s is 2*t1. The distance travelled at the uniform speed is 2*t1*t2.
This gives us 2*t1*t1/ 2 + 2*t1*t2 + 2*t1*t1/ 2 = 32.
Now 2*t1+ t2 = 10
=> t1 = (10 - t2)/ 2
We substitute this in 2*t1*t1/ 2 + 2*t1*t2 + 2*t1*t1/ 2 = 32
=> 2*t1*t1 + 2*t1*t2 = 32
=> 2*[(10 - t2)/ 2] ^2 + 2*[(10 - t2)/ 2]*t2 = 32
=> [(10 - t2)/ 2] ^2 + [(10 - t2)/ 2]*t2 = 16
=> (10 - t2) ^2 + 2*10*t2 - 2*t2*t2 = 64
=> 100 + t2^2 - 2*10*t2 + 2*10*t2 - 2*t2*t2 = 64
=> 100 + t2^2 - 2*t2*t2 = 64
=> t2^2 = 36
=> t2= 6 or -6
As time cannot be negative, so we have t2= 6 s
Therefore the car moves uniformly for 6 s
This question is part of UPSC exam. View all NEET courses
Most Upvoted Answer
A car start moving along a line first with acceleration a=2 starting f...
Given information:
- Initial acceleration (a) = 2 m/s^2
- Initial velocity (u) = 0 m/s
- Time of motion (t) = 10 s
- Average speed (v_avg) = 3.2 m/s
Calculating distance during acceleration:
To calculate the distance covered during acceleration, we can use the formula:
v = u + at,
where v is the final velocity.
Since the car starts from rest, the initial velocity (u) is 0. Plugging in the given values, we have:
v = 0 + 2 * t,
v = 2t.
Now, to find the distance covered during acceleration, we use the formula:
s = ut + (1/2)at^2.
Using the given values, we can substitute the expressions for u and v:
s = 0 * t + (1/2) * 2 * t^2,
s = t^2.
Calculating distance during deceleration:
Since the car comes to rest after deceleration, the final velocity (v) is 0. Using the formula:
v = u + at,
we can rearrange it to find the time taken during deceleration:
t = (v - u) / a,
t = (0 - v) / a,
t = -v / a.
Thus, the time taken during deceleration is -v / a.
Using the given value of acceleration (a = 2 m/s^2) and average speed (v_avg = 3.2 m/s), we can substitute these values into the equation:
t = -3.2 / 2,
t = -1.6 s.
Calculating distance during uniform motion:
We know that the total time of motion is 10 seconds and the time taken during acceleration and deceleration is 1.6 seconds.
Therefore, the time taken during uniform motion can be calculated by subtracting the time taken during acceleration and deceleration from the total time:
t_uniform = t - t_acceleration - t_deceleration,
t_uniform = 10 - 1.6 - 1.6,
t_uniform = 6.8 s.
Calculating distance during uniform motion:
To find the distance covered during uniform motion, we can use the formula:
v_avg = s_uniform / t_uniform.
Rearranging the formula to solve for distance (s_uniform), we have:
s_uniform = v_avg * t_uniform,
s_uniform = 3.2 * 6.8,
s_uniform = 21.76 m.
Conclusion:
The car moves uniformly for a duration of 6.8 seconds and covers a distance of 21.76 meters during that time.
- Initial acceleration (a) = 2 m/s^2
- Initial velocity (u) = 0 m/s
- Time of motion (t) = 10 s
- Average speed (v_avg) = 3.2 m/s
Calculating distance during acceleration:
To calculate the distance covered during acceleration, we can use the formula:
v = u + at,
where v is the final velocity.
Since the car starts from rest, the initial velocity (u) is 0. Plugging in the given values, we have:
v = 0 + 2 * t,
v = 2t.
Now, to find the distance covered during acceleration, we use the formula:
s = ut + (1/2)at^2.
Using the given values, we can substitute the expressions for u and v:
s = 0 * t + (1/2) * 2 * t^2,
s = t^2.
Calculating distance during deceleration:
Since the car comes to rest after deceleration, the final velocity (v) is 0. Using the formula:
v = u + at,
we can rearrange it to find the time taken during deceleration:
t = (v - u) / a,
t = (0 - v) / a,
t = -v / a.
Thus, the time taken during deceleration is -v / a.
Using the given value of acceleration (a = 2 m/s^2) and average speed (v_avg = 3.2 m/s), we can substitute these values into the equation:
t = -3.2 / 2,
t = -1.6 s.
Calculating distance during uniform motion:
We know that the total time of motion is 10 seconds and the time taken during acceleration and deceleration is 1.6 seconds.
Therefore, the time taken during uniform motion can be calculated by subtracting the time taken during acceleration and deceleration from the total time:
t_uniform = t - t_acceleration - t_deceleration,
t_uniform = 10 - 1.6 - 1.6,
t_uniform = 6.8 s.
Calculating distance during uniform motion:
To find the distance covered during uniform motion, we can use the formula:
v_avg = s_uniform / t_uniform.
Rearranging the formula to solve for distance (s_uniform), we have:
s_uniform = v_avg * t_uniform,
s_uniform = 3.2 * 6.8,
s_uniform = 21.76 m.
Conclusion:
The car moves uniformly for a duration of 6.8 seconds and covers a distance of 21.76 meters during that time.
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A car start moving along a line first with acceleration a=2 starting from rest then uniformly and finally decelerating at the same rate and comes to rest the total time of motion is 10s the average speed during the time is 3.2m/s how long does the car moved uniformly?
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A car start moving along a line first with acceleration a=2 starting from rest then uniformly and finally decelerating at the same rate and comes to rest the total time of motion is 10s the average speed during the time is 3.2m/s how long does the car moved uniformly? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A car start moving along a line first with acceleration a=2 starting from rest then uniformly and finally decelerating at the same rate and comes to rest the total time of motion is 10s the average speed during the time is 3.2m/s how long does the car moved uniformly? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A car start moving along a line first with acceleration a=2 starting from rest then uniformly and finally decelerating at the same rate and comes to rest the total time of motion is 10s the average speed during the time is 3.2m/s how long does the car moved uniformly?.
A car start moving along a line first with acceleration a=2 starting from rest then uniformly and finally decelerating at the same rate and comes to rest the total time of motion is 10s the average speed during the time is 3.2m/s how long does the car moved uniformly? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A car start moving along a line first with acceleration a=2 starting from rest then uniformly and finally decelerating at the same rate and comes to rest the total time of motion is 10s the average speed during the time is 3.2m/s how long does the car moved uniformly? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A car start moving along a line first with acceleration a=2 starting from rest then uniformly and finally decelerating at the same rate and comes to rest the total time of motion is 10s the average speed during the time is 3.2m/s how long does the car moved uniformly?.
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