A sprinter starts from rest accelerates to the right or...

A sprinter starts from rest, accelerates to the right or finish line at +2 m/s2 (acceleration vector points toward the right or finish line) for the first 5 seconds, then holds a constant velocity for 3 seconds and finally decelerates at -0.5 m/s2 (deceleration vector points towards the starting block) until the sprinter's velocity is +5 m/s (velocity vector still points toward the right or finish line).that the velocity and displacement values are both equal to 0. a) Sketch the acceleration versus time graph, and from this, the velocity versus time graph to determine the time at which the sprinter has decelerated to +5 m/s. b) Within the first 5 seconds, at what time is the sprinter's velocity +5 m/s (vector pointing towards the right or finish line)?

Transcript

00:01 Hello, it is given in the question that the sprinter starts from rest, that is initial velocity v1 equal to 0 meter per second, and accelerates with an acceleration, a1 equal to 2 meter per second square for time t2, or t1, equal 5 seconds.

00:19 And then attains a constant velocity, let us consider the constant velocity is v2 for time t2 equal to 3 seconds.

00:28 And then the accelerates with an acceleration a3 equal to minus 0 .5 p.

00:33 Meter per second square.

00:35 Here the negative sign is because of the representation of the acceleration here to attain a final velocity v3 which is equal to 5 meter per second and let us consider this happens in time t3.

00:46 So in order to draw the acceleration versus time graph or the velocity versus time graph we need to find v2 and t3.

00:53 So v2 and t3 we need to find what is the value of v2 and t3.

00:57 So let us use the equation of motion v equal to u plus 80.

01:01 So we can write otherwise as b2 and v2 equals v1 plus a1 t1.

01:08 Putting the values of the quantity into the given expression of we can get the value of v2 from here and value of v2 equals 2 multiplied 5 and that equals 10 meter per second.

01:21 And to find the value of t3 we can use the same equation of motion here as we have used v equal to u plus 80 and so we can write v3 equals v2 plus 83 t3.

01:32 So we can put the value of all the acceleration.

01:35 And the velocity here and we can find the value of time t3 so we can put 5 equals 10 minus 0 .5 t3 and this implies that t3 equals 10 seconds so the value of t3 equal 10 seconds now let us draw the acceleration versus time graph so this is the acceleration axis over here and this is the time axis over here so let us consider that the time t is in seconds and the acceleration a is in meter per second square so initially the acceleration is 2 meter per second square for about 5 seconds.

02:11 So this is 5 second and after that the sprinter attains a constant velocity of 10 meter per second so the acceleration is go to 0 for about 3 seconds.

02:18 So this is 8.

02:20 Up to 8 second the sprinter is moving with a constant velocity.

02:23 Then after 8 second the sprinter again moves with the de -acceleration of minus 0 .5 to up to 10 second after that it attains again a constant velocity of 5 meter per second...

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