Respuesta :
Answer:
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
[tex]v_f = v_i + at[/tex]
here we have
[tex]0 = v - at[/tex]
[tex]t = \frac{v}{a}[/tex]
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
[tex]v_f^2 - v_i^2 = 2 ad[/tex]
[tex]d = \frac{0 - v^2}{2(-a)}[/tex]
[tex]d = \frac{v^2}{2a}[/tex]
since the speed is double so the distance to stop the car will becomes 4 times
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:
[tex]a(t) = -A[/tex]
Where the negative sign is because this is a deceleration.
To get the velocity equation we integrate over time to get:
[tex]v(t) = -A*t + v_0[/tex]
Where v₀ is the initial velocity at which the car was moving.
To get the position equation we integrate again to get:
[tex]p(t) = -(A/2)*t^2 + v_0*t[/tex]
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:
[tex]v(t) = 0 = -A*t + v_0[/tex]
[tex]t = v_0/A[/tex]
Now if you double the velocity (the initial velocity, 2*v₀)
Then the velocity equation becomes:
[tex]v'(t) = -A*t + 2*v_0[/tex]
And the time needed to come to a full stop is:
[tex]v'(t) = 0 = -A*t + 2*v_0[/tex]
[tex]t = 2*v_0/A[/tex]
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:
[tex]p(v_0/A) =-(A/2)*(v_0/A)^2 + v_0*v_0/A = (1/2)*v_0^2/A[/tex]
While for the case where the velocity is doubled, the position equation becomes:
[tex]p'(t) = -(A/2)*t^2 + 2*v_0*t[/tex]
And the time in this case is:
[tex]t = 2*v_0/A[/tex]
Evaluating the function in this time we get:
[tex]p'(2*v_0/A) = -(A/2)*(2*v_0/A)^2 + 2*v_0*(2*v_0/A)[/tex]
[tex]= 4*( -(A/2)*(v_0/A)^2 + v_0*v_0/A ) = 2*v_0^2/A[/tex]
So the distance needed to stop is 4 times larger when you double the velocity.
If you want to learn more, you can read:
https://brainly.com/question/23774048
