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A car starts moving along a line,first with accelaration a=2m/s^2,starting from rest...

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andyluv | Student, Grade 11 | Honors

Posted November 6, 2010 at 9:09 PM via web

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A car starts moving along a line,first with accelaration a=2m/s^2,starting from rest then uniformally moving and finally deaccelarating at the same

rate and comes to rest.The total time of motion is10 seconds.The average speed during the time is 3.2 m/s.How long does the car move uniformly?

Tagged with kinematics, physics, science

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sociality | High School Teacher | Valedictorian

Posted November 7, 2010 at 3:41 AM (Answer #1)

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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

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neela | High School Teacher | Valedictorian

Posted November 7, 2010 at 4:10 PM (Answer #2)

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The product average velocity and time is the distance travelled by the car = 3.2m/s *10s = 32 m.

The initial velocity of the car = u = 0, uniform acceleration a = 2m/s^2. Then the final velocity = u+at = 0+2t = t.The deceleration being 2m/s^2, the time required to attain the zero velocity from  the initial velocity of  2t is t. Threfore the time of uniform running of the car = 10-2t seconds.

The distance travelled by the car from rest  or the initial velocity of 0 m/s  to reach the final velocity  of 2t with a uniform acceleration of 2m/s^2 = (initial speed+final speed)/2 * time = 0+2t)t= t^2.

The car travels now with uniform speed of 2t m/s for (10-2t) secs. So the distance travelled = 2t(10-2t = 20t-4t^2.

The car now decelerates uniformly at 2m/s^2 and from the initial speed of 2tm/2 till its speed is zero, for this tha car takes t time. Therfore the distance travelled by the car in this time = (initial velocity+final velocity)/2 * time = {(2t+0)/2}*t = t^2.

Thus the distance travelled by the car = t^2+20t-4t^2 +t^2 which should be 32 meter.

Therefore 20t-2t^2 = 32. Or 2t^2-20t+32= 0. Or dividing this equation by 2 , we get: t^-10t +16 = 0. Or (t-8)(t-2) = 0 Therefore t= 2 is the  possible time.

The time during  which the car travels with urniform speed is 10-2t = 10-2*2 = 6 seconds. The uniform speed of the car = 2t = 2*2 = 4m/s. Distance travelled with uniform speed = 4*6 = 32 meter.

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