In a women's 100-m race, accelerating uniformly, Laura takes 2.00 s and Healan 3.00 s to attain their maximum speeds, which they each maintain for the rest of the race. They cross the finish line simultane- ously, both setting a world record of 10.4 s

Complete question is;

In a women’s 100-m race, accelerating uniformly, Laura takes 2.00 s and Healan 3.00 s to attain their maximum 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 s.

(a) What is the acceleration of each sprinter?

(b) What are their respective maximum speeds?

(c) Which sprinter is ahead at the 6.00-s mark, and by how much?

(d) What is the maximum distance by which Healan is behind Laura, and at what time does that occur?

A) a_laura = 5.319 m/s² and a_healan = 3.745m/s²

B) V_max for laura = 10.638 m/s and V_max for Healan = 11.235 m/s

C) Laura is ahead of healan by 2.632 m

D) Healan is 4.471 m behind Laura

Explanation:

A) The distance traveled by each sprinter with constant acceleration would be gotten from the equation of motion which is;

s = ut + ½at²

Now, since they began at rest, thus, initial velocity(u) = 0 m/s

So, s = 0 + ½at²

s = ½at²

Now, maximum speed achieved by each of them is given by the formula;

v_max = u + at

Since u = 0, then;

v_max = at

Now, we are told that they set a world record of 10.4 s, thus;

Time taken by each to run at their maximum speed = (10.4 - t)

Distance travelled after attaining maximum speed = at + a(10.4 - t)

Thus;

Total distance covered by each sprinter = ½at² + (at)(10.4 - t)

Where;

t = time taken to accelerate to maximum speed

Since total distance covered is 100 m and she took 2 s to accelerate to max speed, then Acceleration for Laura would be calculated from;

100 = ½a(2)² + (2a(10.4 - 2))

This gives;

100 = 2a + 16.8a

18.8a = 100

a = 100/18.8

a_laura = 5.319 m/s²

Similarly, for Healan:

100 = ½a(3²) + (3a(10.4 - 3)

100 = 4.5 a + 22.2a

26.7a = 100

a = 100/26.7

a_healan = 3.745m/s²

B) From earlier we saw that max speed is;

v_max = at

Thus, for laura;

Vmax for laura= 5.319 × 2

V_max for laura = 10.638 m/s

Also, for healan;

Vmax = 3.745 × 3

V_max for Healan = 11.235 m/s

C) to find which sprinter is ahead at 6.00 s,

For Laura:

D = (½(5.319 × 2²)) + (2² × 10.638)

D_laura = 53.19 m

For Helen: D = (½(3.745 × 3²)) + (3 × 11.235) = 50.558 m

Difference = 53.19 - 50.558 = 2.632

Thus, Laura is ahead by 2.632 m

C) We are told that laura achieved max speed in 2 s while healan took 3 s. This means that when Laura had reached her maximum speed Healan was still accelerating. Thus;

From v_max = at

10.638 = (3.745)t

t = 10.638/3.745

t = 2.84 s

Distance travelled during this time is;

For Laura:

D = (½(5.319 × 2²) + (0.84)(10.638) = 19.574 m

Foe Healan:

D = (½(3.745 × 2.84²)) = 15.103 m

Difference = 19.574 - 15.103 = 4.471 m

Thus, Healan is 4.471 m behind laura

the velocity you are looking for is made up of two parts: horizontally velocity, vh, and its vertical velocity, vv. the actual velocity of the ball will be these two parts added together, and because velocity is a vector then they will need to be added vectorially.  

neglecting air resistance etc etc.,  

the horizontal velocity component will be constant (because there is no accelerating force in the horizontal direction.  

vh = 8 ms-1  

the vertical velocity component has an accelerating force: 'g'.  

using v = u +at where a = g = 9.81 ms-2.  

the vertical component of velocity, vv = 0 + gt = vv =9.81 x 2  

vv = 19.62 ms-1 . the zero appears because the virtical component of velocity at time t=0 is zero. remember, the ball was thrown horizontally.  

how fast it will be moving (it actual path (being a parabola )) will be the vector sum of these two.  

adding vectorally gives: -  

v = sqr root (8^2 + 19.62^2) = sqr root(64+384.9) = 21.18 or 21.2 ms-1

the first, second, and fifth options