A runner is taking an 800 -m race. He accelerates uniformly from start for the f

Question

A runner is taking an 800 -m race. He accelerates uniformly from start for the first 160 m and then runs with a constant velocity, until the last 120 m. The sprinter then accelerates during this last 120 m, at the same rate ho did In the first 160 m of the race. to finish the race. If the sprinter's time for the first 160 m is 30 seconds, determine his acceleration during the first 30 seconds his velocity before the last 120 meters his final velocity at the end of the race the total time he takes to complete the 800 - m race.

Explanation / Answer

For all the subdivisions we are going to use the three formulas given below:

v = u+at

s = ut + at^2/2

v^2 = u^2 +2as

u = initial velocity, v = final velocity, s = distance covered, t = time taken

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a) s = 160 m, t=30, u = initial velocity =0

Hence use the second formula

s = ut + at^2/2

160 =0+a(900)/2

a = 320/900 = 16/45 m sec-2

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b) Velocity before 120 m is the velocity after 160 m (since constant velocity after 160th m)

So use third formula

v^2 = u^2 +2as.

u=0 s=160 and a = 16/45

So v^2 = 0+32/45 * 160

v = 32/3 m/sec

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c) Let velocity at the end of race be v1.

v1^2 = (32/3)^2+2(16/45) 120

since 32/3 is initial velocity and a =16/45 and s = 120

Simplify to have

v = 14.11067 m/sec

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d) Total time to complete the race

= t1+t2+t3

t1 = time taken to complete 160 m = 30 s

t2 = time taken to complete (680-160) m = 520/(32/3)= 48.75 s

t3 = time taken for last 120 m

14.11067= 32/3 + 16t/45

t3 = 9.686

Total time taken = 78.75+9.686

= 88.436 s.

  

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