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1 year ago

8

give an example of situation in which an automobile driver can have a centripetal acceleration but no tangential speed

1 answer:

Mars2501 [29]1 year ago

6 0

There is no need for tangential acceleration when moving in a circle at a constant speed.

<h3>What is centripetal acceleration?</h3>

centripetal acceleration refers to the speed at which a body moves through a circle. Due to the fact that velocity is a vector quantity (i.e., it has both a magnitude, the speed, and a direction), when a body travels in a circle, its direction is constantly changing, which causes a change in velocity, which results in an acceleration.

<h3>Which is an example of centripetal acceleration?</h3>

Centripetal acceleration occurs when you spin a ball on a string above your head. A car experiences centripetal acceleration when it is being driven in a circle. Additionally, a satellite in orbit around the Earth experiences centripetal acceleration.

To know more about tangential acceleration :

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3 years ago

A car traveling on a flat (unbanked) circular track accelerates uniformly from rest with a tangential acceleration of 1.7 m/s^2.

Komok [63]

Answer:

The coefficient of static friction between the car and the track

u=0.572

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∑ $F=F_{f}=m*a_{t}$

As we know

$F_{f}=u*F_{N}=u*m*g$

Also the center ward direction forces

$F_{fc}=m*a_{c}$

$a_{c}=\frac{v_{t}^2}{r}$

$F_{fc}=m*\frac{v_{t}^2}{r}$

But now vt relation with the tangential acceleration

$v_{t}=2*a_{t}*\frac{\pi }{r}$

replacing

$F_{fc}=m*a_{t}*\frac{2\pi*r}{2r}$

$F_{fc}=m*a_{t}*\pi$

So magnitude of the force can get by

$F_{f}=\sqrt{(m*a_{t}*\pi)^{2}+(m*a_{t})^{2}}$

Get the factor to simplify

$F_{f}=a_{t}*m*\sqrt{(1+\pi^2)}$

$u*m*g=m*a_{t}*(\sqrt{1+\pi^2})$

Solve to u'

$u=\frac{a_{t}}{g}*(\sqrt{1+\pi^2})$

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3 years ago

Derive the formula v= u+ at<br>please derive in detail

gogolik [260]

Lets do

We know

The rate of change of velocity is acceleration .

$\\ \sf\longmapsto a=\dfrac{dv}{dt}$

$\\ \sf\longmapsto dv=adt$

Integrate both sides

$\\ \sf\longmapsto \int dv=a\int dt$

As acceleration is constant .Take it outside of integral .On velocity we can take limit u to v and time from 0 to t

$\\ \sf\longmapsto {\displaystyle{\int}}^v_u dv=a{\displaystyle{\int}}^t_0 dt$

Hence

$\\ \sf\longmapsto v{\huge{|}}^v_u=at$

$\\ \sf\longmapsto v-u=at$

$\\ \sf\longmapsto v=u+at$

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3 years ago

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