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skad [1K]
3 years ago
6

An Indy 500 race car's velocity increases from 4.0 m/s to +36

Physics
1 answer:
Soloha48 [4]3 years ago
3 0

Answer:

The average acceleration is 8\ m/s^2

Explanation:

<u>Uniform Acceleration </u>

When an object varies its velocity at the same rate, the acceleration is constant.

The relation between the initial and final speeds is:

v_f=v_o+a.t

Where:

vf  = Final speed

vo = Initial speed

a   = Constant acceleration

t   = Elapsed time

The acceleration can be calculated by solving for a:

\displaystyle a=\frac{v_f-v_o}{t}

The Indy 500 race car increases its speed from vo=4 m/s to vf=36 m/s in t=4 s. Thus, the average acceleration is:

\displaystyle a=\frac{36-4}{4}=8\ m/s^2

The average acceleration is 8\ m/s^2

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A ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.665 m. W
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Answer:

Impulse is 1.239 kg.m/s in upward direction

Explanation:

Taking upward motion as positive and downward motion as negative.

Downward motion:

Given:

Mass of ball (m) = 0.150 kg

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Initial velocity (u) = 0 m/s

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Using equation of motion, we have:

v_d^2=u^2+2aS\\\\v=\pm\sqrt{u^2+2aS}\\\\v_d=\pm\sqrt{0+2\times -9.8\times -1.25}\\\\v_d=\pm\sqrt{24.5}=\pm4.95\ m/s

Since, the motion is downward, final velocity must be negative. So,

v_d=-4.95\ m/s

Upward motion:

Given:

Displacement of ball (S) = 0.665 m

Initial velocity (v_d) = 4.95 m/s(Upward direction)

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

v_{up}^2=v_d^2+2aS\\\\v_{up}=\pm\sqrt{v_d^2+2aS}\\\\v_{up}=\pm\sqrt{24.5+2\times -9.8\times 0.665}\\\\v_{up}=\pm\sqrt{10.966}=\pm3.31\ m/s

Since, the motion is upward, final velocity must be positive. So,

v_{up}=3.31\ m/s

Now, impulse is equal to change in momentum. So,

Impulse = Final momentum - Initial momentum

J=m(v_{up}-v_d)\\\\J=(0.150\ kg)(3.31-(-4.95))\ m/s\\\\J=0.150\ kg\times 8.26\ m/s\\\\J=1.239\ Ns

Therefore, the impulse given to the ball by the floor is 1.239 kg.m/s in upward direction.

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