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kap26
3 months ago
11

A motorcycle traveling at 36 m/s slams on the brakes to avoid an accident. The motorcycle skids 23m before stoping. What is the

motorcycle’s acceleration? How long does it take to come to a stop
Physics
1 answer:
ValentinkaMS [3.4K]3 months ago
8 0
Since the motorcycle was at a speed of 36 m/s prior to braking, that marks the initial velocity.
u = 36 {ms}^{ - 1}
The motorcycle skidded 23m before coming to a full stop, indicating that
s = 23m
As it has ceased motion, the final velocity is zero.

v = 0{ms}^{ - 1}
We can apply the 'suvat' formula relevant to linear motion.

{v}^{2} = {u}^{2} + 2as
Substituting the aforementioned values allows us to find,

{0}^{2} = {36}^{2} + 2a(23)


0 = 1296+ 46a
46a = - 1296
a = - 28.2 {ms}^{ - 2}


We can apply the formula
v = u + at
to calculate the time required for the motorcycle to stop.

0 = 36 + - 28.2t
- 36 = - 28.2t
t = 1.3s
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At great distances from the planet, if neglecting other gravitational influences, the rock will reach a speed close to \sqrt{3} v_{e}

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The initial velocity of the rock, v_{i}=2 v_{e}. Given that the radius is R, we can determine the escape velocity using the following formula:

            \frac{1}{2} m v_{e}^{2}-\frac{G M m}{R}=0

            \frac{1}{2} m v_{e}^{2}=\frac{G M m}{R}

            v_{e}^{2}=\frac{2 G M}{R}

            v_{e}=\sqrt{\frac{2 G M}{R}}

Where M represents the mass of the planet and G represents the gravitational constant.

Based on the given conditions,

Surface potential energy can be articulated as,  U_{i}=-\frac{G M m}{R}

As R approaches infinity when distanced from the planet, thus v_{f}=0

Then, the initialkinetic energy will be expressed as

                  k_{i}=\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m\left(2 v_{e}\right)^{2}

Simultaneously, the finalkinetic energy can be described as

k_{f}=\frac{1}{2} m v_{f}^{2}                

v_{f}=\text { final velocity }Here,

Now, adding the potential and kinetic energies at the beginning and end gives the equation to determine the final velocity

expressed as

U_{i}+k_{i}=k_{f}+v_{f}                  

\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}+0                  

\frac{1}{2} m\left(2 v_{e}\right)^{2}-\frac{G M m}{R}=\frac{1}{2} m v_{f}^{2}                  

\frac{1}{2}By canceling 'm' common on both sides, we yield

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Recognizing that v_{e}=\sqrt{\frac{2 G M}{R}}, we can rewrite it as \left(v_{e}\right)^{2}=\frac{2 G M}{R}, we arrive at

                4\left(v_{e}\right)^{2}-\left(v_{e}\right)^{2}=v_{f}^{2}

                v_{f}^{2}=3\left(v_{e}\right)^{2}

Taking the squares out leads to

                v_{f}=\sqrt{3} v_{e}

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