Question

In a woman's 100-m race, accelerating uniformly, Laura takes 2.00 s and Healan 3.00 s to attain thier maxiumum speeds, which they each maintain for the rest of the race. They cross the finish line simultaneously, both setting a world record of 10.4 second a) what is the acceleration of each sprinter? b) What are thier respective maximum speeds ? c) Which sprinter is ahead at the 6.00-s mark, and) what is the maximum distance by which Healan is behind Laura , and at what time does that occur?

Answer 1

Suppose acceleration of Helen is a1 and acceleration of laura is a2, then

Using 1st kinematic equation:

V1 = U1 + a1*t1

V1 = Speed of helen after 2.00 sec

U1 = initial speed of helen = 0 m/sec

t1 = 2.00 sec, which gives

V1 = 0 + a1*2.00

V1 = 2*a1 = max speed of Helen

Similar max speed of Laura will be:

V2 = U2 + a2*t2

V2 = 0 + a2*3.00

V2 = 3*a2 = max speed of Laura

Now Since they both finishes simultaneously, So total distance traveled by both of them is 100 m in 10.4 sec

Now for Helen

D = d1 + d2

d1 = distance traveled by helen during sprinting = U1*t1 + (1/2)*a1*t1^2

d1 = 0*2 + (1/2)*a1*2^2 = 2*a1

d2 = distance traveled by Helen at max constant speed = V1*t1'

t1' = 10.4 - 2.0 = 8.4 sec

d2 = V1*t1' = 2*a1*8.4 = 16.8*a1

Since d1 + d2 = 100 m, So

2*a1 + 16.8*a1 = 100

a1 = 100/18.8 = 5.32 m/sec^2 = acceleration of Helen

Now for Laura

D = d1 + d2

d1 = distance traveled by Laura during sprinting = U2*t2 + (1/2)*a2*t2^2

d1 = 0*3 + (1/2)*a2*3^2 = 4.5*a2

d2 = distance traveled by Laura at max constant speed = V2*t2'

t2' = 10.4 - 3.0 = 7.4 sec

d2 = V2*t2' = 3*a2*7.4 = 22.2*a2

Since d1 + d2 = 100 m, So

4.5*a2 + 22.2*a2 = 100

a2 = 100/26.7 = 3.745 m/sec^2 = acceleration of Laura

Part B.

Max speed of Helen will be:

V1 = 2*a1 = 2*5.32

V1 = 10.64 m/sec

Max speed of Laura will be:

V2 = 3*a2 = 3*3.745

V2 = 11.235 m/sec

Part C.

At t = 6.00 sec, distance traveled by Helen will be:

d1 = distance traveled by Helen in 2 sec = 2*a1 = 2*5.32 = 10.64 m

d2 = distance traveled by Helen in next 4 sec = V1*T = 10.64*4 = 42.56 m

Total distance traveled by Helen = 10.64 + 42.56 = 53.2 m

At t = 6.00 sec, distance traveled by Laura will be:

d1 = distance traveled by Laura in 3 sec = 4.5*a2 = 4.5*3.745 = 16.85 m

d2 = distance traveled by Laura in next 3 sec = V2*T = 11.235*3 = 33.705 m

Total distance traveled by Laura = 16.85 + 33.705 = 50.55 m

So from above we can see that at 6.00 sec mark Helen will be ahead of Laura

Part D.

Since Helen has more acceleration than Laura, So distance between them will be max before Laura reach his top speed, which means max distance between them will be between 2 and 3 sec, Now to find that distance suppose distance between then at time 't' (2 < t < 3) is max and at this time distance between them is D, then

D = D1 - D2

D1 = distance traveled by Helen = d1 + d2 = 2*a1 + V1*t

D2 = distance traveled by Laura = U1*t + (1/2)*a2*t^2 = 0*t + (1/2)*a2*t^2

So,

D = 2*a1 + V1*(t - 2) - (1/2)*a2*t^2

D = -(1/2)*a2*t^2 + V1*t + (2*a1 - 2*V1)

to find max D, differentiate above equation

D' = -a2*t + V1 + 0 = 0

t = V1/a2 = 10.64/3.745

t = 2.84 sec = time when distance between Helen and Laura will be maximum

D = -(1/2)*a2*t^2 + V1*t + (2*a1 - 2*V1)

D = -(1/2)*3.745*2.84^2 + 10.64*2.84 + (2*5.32 - 2*10.64)

D = 4.47 m = max distance between Helen and Laura

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