A certain sprinter has a top speed of 10.6 m/s. If the sprinter starts from rest and accelerates at a constant rate, he is able to reach his top speed in a distance of 11.9 m. He is then able to maintain his top speed for the remainder of a 100 m race. (a) What is his time for the 100 m race? (b) In order to improve his time, the sprinter tries to decrease the distance required for him to reach his top speed. What must this distance be if he is to achieve a time of 10.4 s for the race?

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rejkjavik rejkjavik · Answer 1 · 2019-09-11T12:35:54+02:00

Answer:

total time is= 9.64 s

[tex]\Delta x1 = 10.24 m[/tex]

Explanation:

average velocity = 1/2(vinitial + vfinal)

average velocity = 5.3m/s,

and therefore it took a time of

time = distance/velocity

= 11.9m/5.3m/s

= 2.24 s

he ran the remaining 88.1 m at a speed of 11.9 m/s, or in a time of 7.40s

his total time is= 2.24s+ 7.40s = 9.64 sec

b)

for constant acceleration stage, the distance is given as

[tex]\Delta x1 =\frac{u +v)t1}{2}[/tex]

[tex] =\frac{0+10.6}t1}{2}[/tex]

= 5.3t1

for constant speed

[tex]\Delta x2 =ut2[/tex]

[tex]100 - \Deltax1 =10.6 (10.4 -t1)[/tex]

[tex]100 - \Deltax1 = 110.24 - 10.6 t1[/tex]

we know that[tex] \Delta x1 = 5.3t2[/tex], so

100 - 5.3t1 = 110.24 - 10.6 t1

solving for t1

t1 = 1.93 sec

therefore

[tex]\Delta x1 = 5.3*1.93 = 10.24[/tex]

[tex]\Delta x1 = 10.24 m[/tex]