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3 years ago

How do you calculate the braking distance

Physics

1 answer:

Anettt [7]3 years ago

6 0

The braking distance is given by $s=\frac{-u^2}{2a}$

Explanation:

When the driver of a car hits the pedal of the brakes, the car starts decelerating until it stops. Assuming the deceleration is constant, then the motion is a uniformly accelerated motion, so we can use the following suvat equation:

$v^2-u^2=2as$

where

u is the initial speed of the car

v is the final speed of the car, which is zero because the car comes to rest:

v = 0

a is the acceleration of the car

s is the distance travelled by the car during the deceleration, so it is the braking distance

Therefore, re-arranging the equation for s, we find an expression for the braking distance:

$s=\frac{-u^2}{2a}$

Note that the sign of $a$ is negative since the car is decelerating, therefore the final sign of $s$ is positive.

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

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Answer:

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Explanation:

From the question we are told that

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The time is $t = 0.15s$

The velocity at t is $v = 16 m/s$

Generally average angular acceleration is mathematically represented as

$\alpha = \frac{w_f - w_o}{t}$

Where $w_f$ is the finial angular velocity which is mathematically evaluated as

$w_f = \frac{v}{l}$

$w_f = \frac{16}{0.85}$

$= 18.823 rad/s$

and $w_o$ is the initial angular velocity which is zero since initial linear velocity is zero

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3 years ago

During a tennis volley, a ball that arrives at a player at 40 m/s is struck by the racquet and returned at 40 m/s. The other pla

Butoxors [25]

Answer:Racquet force is twice of Player force

Explanation:

Given

ball arrives at a speed of $u=-40\ m/s$

ball returned with speed of $v=40\ m/s$

average Force imparted by racquet on the ball is given by

$F_{racquet}=\frac{m(v-u)}{\Delta t}$

where $m=mass\ of\ ball$

$\Delta t=$ time of contact of ball with racquet

$F_{racquet}=\frac{m(40-(-40))}{\Delta t}$

$F_{racquet}=\frac{80m}{\Delta t}-----1$

When it land on the player hand its final velocity becomes zero and time of contact is same as of racquet

$F_{player}=\frac{m(0-40)}{\Delta t}$

$F_{player}=\frac{-40m}{\Delta t}-----2$

From 1 and 2 we get

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3 years ago

A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh

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Answer:

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Explanation:

From the question we are told that

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