Physics, 17.07.2019 11:30, franciscovazquez623

Assume that the brakes in your car create a constant deceleration regardless of how fast you are going. if you double your driving speed, how does this affect (a) the time required to come to a stop, and (b) the distance needed to stop?

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answered: jolayemihazeez97

Deceleration is defined as an acceleration in the opposite way of the motion direction.

We will find that the answers are:

a) The time required to come to a stop is doubled

b) The distance required to come to a stop is quadrupled.

Let's write the motion equations for the given situation.

If you decelerate at a constant rate, the acceleration will be given by:

a(t) = -A

Where the negative sign is because this is a deceleration.

To get the velocity equation we integrate over time to get:

v(t) = -A*t + v_0

Where v₀ is the initial velocity at which the car was moving.

To get the position equation we integrate again to get:

p(t) = -(A/2)*t^2 + v_0*t

a) With these equations, the time required to come to a stop is the value of t such that the velocity becomes equal to zero.

Then we will get:

v(t) = 0 = -A*t + v_0

t = v_0/A

Now if you double the velocity (the initial velocity, 2*v₀)

Then the velocity equation becomes:

v'(t) = -A*t + 2*v_0

And the time needed to come to a full stop is:

v'(t) = 0 = -A*t + 2*v_0

t = 2*v_0/A

So the time to come to a full stop is twice the one we got before.

b) The distance needed to stop comes from evaluating the position equation in the time we found earlier.

For the first case we get:

p(v_0/A) =-(A/2)*(v_0/A)^2 + v_0*v_0/A = (1/2)*v_0^2/A

While for the case where the velocity is doubled, the position equation becomes:

p'(t) = -(A/2)*t^2 + 2*v_0*t

And the time in this case is:

t = 2*v_0/A

Evaluating the function in this time we get:

p'(2*v_0/A) = -(A/2)*(2*v_0/A)^2 + 2*v_0*(2*v_0/A)

= 4*( -(A/2)*(v_0/A)^2 + v_0*v_0/A ) = 2*v_0^2/A

So the distance needed to stop is 4 times larger when you double the velocity.

If you want to learn more, you can read:

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answered: lilquongohard

Part a)

Time becomes double to stop it

Part b)

Distance to stop the car will become 4 times

Explanation:

Part a)

To find the time to stop the car we know that final speed is zero

so we will have

v_f = v_i + at

here we have

0 = v - at

$t = \frac{v}{a}$

Now since the speed is double

so the time to stop will also becomes double

Part b)

To find the stopping distance we know that final speed of the object must be zero

so we will have

v_f^2 - v_i^2 = 2 ad

$d = \frac{0 - v^2}{2(-a)}$

$d = \frac{v^2}{2a}$

since the speed is double so the distance to stop the car will becomes 4 times

answered: Guest

The answer is b hope this

answered: Guest

it's when something is going so fast it makes a loud noise, breaking the sound barrier, for example: a jet makes a loud noise, not only cause of the jet but because they reached a point where they went so fast that you can year the noise from various parts, kinda like a police car, lets say the car was going so fast that you hear the siren of the cop car right next to you, but the car is probably a mile a way, it's cos the siren was next to you a few seconds if you understand me

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Assume that the brakes in your car create a constant deceleration regardless of how fast you are goi...