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3 years ago

A) Megan was doing time-trials on her bike around 400 metre horizontal track.

Physics

1 answer:

Georgia [21]3 years ago

8 0

A) The forward force is equal to the backward force

In this problem:

- the forward force is the force that Megan applies to the pedal to go forward

- the backward force is due to the air resistance and the friction between the wheels and the track

In this case, Megan is travelling at constant speed. This means that her acceleration is zero:

a = 0

According to Newton's second law, the resultant of the forces acting on Megan is equal to the product between mass (m) and acceleration (a):

$\sum F = ma$

However, a = 0, so the resultant of the forces is also zero:

$\sum F =0$

and this implies that the forward force and the backward force are equal in magnitude and opposite in direction.

B) The forward force is larger than the backward force

In this case, Megan crouched down in order to make herself more streamlined. As a result, the air resistance acting on Megan will decrease: so, the backward force will decrease, and therefore the forward force (which has remained the same) will be larger than the backward force.

So, the resultant force

$\sum F$

will be no longer zero, and therefore the acceleration will be different from zero, which means that Megan will increase her speed.

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A man jogs at a speed of 1.6 m/s. His dog waits 1.8 s and then takes off running at a speed of 3 m/s to catch the man. How far w

inessss [21]

Answer:

The dog catches up with the man 6.1714m later.

Explanation:

The first thing to take into account is the speed formula. It is $v=\frac{d}{t}$ , where v is speed, d is distance and t is time. From this formula, we can get the distance formula by finding d, it is $d=v\cdot t$

Now, the distance equation for the man would be:

$d_{man}=v_{man}\cdot t=1.6\cdot t$

The distance equation for the dog would be obtained by the same way with just a little detail. The dog takes off running 1.8s after the man did. So, in the equation we must subtract 1.8 from t.

$d_{dog}=v_{dog}\cdot (t-1.8)=3\cdot (t-1.8)$

For a better understanding, at t=1.8 the dog must be in d=0. Let's verify:

$d_{dog}=v_{dog}\cdot (1.8-1.8)=3\cdot (0)=0$

Now, for finding how far they have each traveled when the dog catches up with the man we must match the equations of each one.

$d_{man}=d_{dog}$

$1.6\cdot t=3\cdot (t-1.8)$

$1.6\cdot t=3\cdot t-5.4$

$1.4\cdot t=5.4$

$t=\frac{5.4}{1.4}$

$t=3.8571s$

The result obtained previously means that the dog catches up with the man 3.8571s after the man started running.

That value is used in the man's distance equation.

$d_{man}=1.6\cdot t=1.6\cdot (3.8571)$

$d_{man}=6.1714m$

Finally, the dog catches up with the man 6.1714m later.

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