Question

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1) Human vs Horse. A human's higher initial acceleration suggests that a human might outrun a horse over a very short race, but the horse's higher top speed allows it win a longer race. A simple-but plausible model for a sprint by a human and a horse uses the following assumptions: The human accelerates at 8.00 m/s for 1.40 s then runs at constant speed. A horse accelerates at a more modest 5.60 m/s for a longer time of 3.40 s and then continues at constant speed. Say a human and a horse are competing in a 200 m race. The human is given a 100 m head start, so he begins 100 m from the finish line a) Draw a diagram of this situation, showing the initial position of both the human and the horse, and the finish line. Draw a single x-axis for both human and horse to describe these positions. Place the origin of your coordinate system at the horse's original position. Eventually, the human and horse will meet; label the meeting point on your diagram as x. Also label the displacement of both human and horse on the diagram, using the meeting point as the final position of both. (2) Assuming that both competitors start from rest, find the top speed of both the human and the horse. Also find their positions when they reach this top speed. (4) (Note: For comparisons' sake, Usain Bolt can achieve a top speed of roughly 12 m/s, while the fastest recorded speed for a racehorse was roughly 19.7 m/s, achieved by Winning Brew in 2008) Find the position x where they meet. (2) Explain how this meeting point shows who wins the race. What value must this position be compared to make this conclusion? (2) b) c) d)

Answer 1

To answer the questions above we find the trajectory of the human and the horse in terms of position and time and then plot the equation on the graph.

we will use newtons laws of motion under constant acceleration to solve our problem.

a)

For human,

since it starts from rest u=0ms^-1

during acceleration.

$x-x_{o}=ut+\frac{1}{2}at^{2}$

$\Rightarrow x-100=0+\frac{1}{2}*8t^{2}$

$\Rightarrow x=4t^{2}+100$..............................(i)

after reaching top speed

$x-x_{o}=ut+\frac{1}{2}at^{2}$

here, x_o=107.84 , u=11.2ms^-1 , t=t-1.4s and a =0

here t is taken as t-1.4 because the first 1.4s were used to accelerate.

$\Rightarrow x-107.84=11.2*(t-1.4)+0$

$\Rightarrow x=11.2t-15.68+107.84$

$\Rightarrow x=11.2t+92.16$........................(ii)

For the horse

since it starts from rest u=0ms^-1

during acceleration.

$x-x_{o}=ut+\frac{1}{2}at^{2}$

$\Rightarrow x-0=0+\frac{1}{2}*5.6t^{2}$

$\Rightarrow x=2.8t^{2}$..........................(iii)

after reaching top speed

$x-x_{o}=ut+\frac{1}{2}at^{2}$

here, x_o=32.368 , u=19.04ms^-1 , t=t-3.4s and a =0

$\Rightarrow x-32.368=19.04*(t-3.4)+0$

$\Rightarrow x=19.04t-64.736+32.368$

$\Rightarrow x=19.04t-32.368$.........................(iv)

rough diagram of the graph

b)

Human :

Top speed after full acceleration

$v=u+at$

$\Rightarrow v=8*1.4ms^{-1}$

$\Rightarrow v=11.2ms^{-1}$

position at which top speed is achieved

$\Rightarrow x=4*1.4^{2}+100$

$\Rightarrow x=7.84+100m$

$\Rightarrow x=107.84m$

Horse:

Top speed after full acceleration

$v=u+at$

$\Rightarrow v=5.6*3.4ms^{-1}$

$\Rightarrow v=19.04ms^{-1}$

position at which top speed is achieved

$\Rightarrow x=2.8*3.4^{2}$

$\Rightarrow x=32.368m$

c)

To find the meeting point.

We can see from the above calculations that just after acceleration the positions does not match. So, if they meet it must be at a point after they have reached top speed.

Now,

to meet there positions must be same so,

equation (ii)=(iv)

$\Rightarrow 11.2t+92.16=19.04t-32.368$

$\Rightarrow 7.84t=124.528$

$\Rightarrow t=\frac{124.528}{7.84}s$

$\Rightarrow t=15.88s$

Therefore, position at which they meet

x=270m

d) The meeting point shows that the race was won by the human. This value must be compared with the distance in which the race occured(i.e 200m)

The meeting point shows that the human and horse meet after they have passed the finish line.

If any doubt feel free to comment.