Question

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?

Answer 1

total time is= 9.64 s

`\Delta x1 = 10.24 m`

Step-by-step 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

`\Delta x1 =(u +v)t1)/(2)`

`=\frac{0+10.6}t1}{2}`

= 5.3t1

for constant speed

`\Delta x2 =ut2`

`100 - \Deltax1 =10.6 (10.4 -t1)`

`100 - \Deltax1  = 110.24 - 10.6 t1`

we know that
`\Delta x1 = 5.3t2`, so

100 - 5.3t1 = 110.24 - 10.6 t1

solving for t1

t1 = 1.93 sec

therefore

`\Delta x1 = 5.3*1.93 = 10.24`

`\Delta x1 = 10.24 m`