Question

A car stops in 150 m. If it has an acceleration of –7.0m/s2, what was the cars starting velocity?

Answer 1

Answer:

We know that:

The car stops in 150 m, so this will be the final position.

The acceleration is -7.0m/s^2

I will assume that the initial position is zero.

Then we can write the movement equations as:

Acceleration:

A(t) = -7.0m/s^2

For the velocity, we integrate over time and get:

V(t) = (-7m/s^2)*t + V0.

Now we can integrate again over the time to get the position.

P(t) = (-3.5 m/s^2)*t^2 + V0*t + P0

Where P0 is the initial position,, ut it is equal to zero.

Then the equations that we need to use are:

V(t) = (-7m/s^2)*t + V0.

P(t) = (-3.5 m/s^2)*t^2 + V0*t

The car will reach a complete stop when:

The velocity is equal to zero

The position is equal to 150m

Then we have a system of equations:

0m/s = (-7m/s^2)*t + V0.

150m = (-3.5 m/s^2)*t^2 + V0*t

To solve this, first let's isolate one variable in one of the equations, i wil isolate V0 in the first equation:

V0 = (7m/s^2)*t

Now i can replace this into the second eqation:

150m = (-3.5m/s^2)*t^2 +  (7m/s^2)*t^2

150m = (3.5m/s^2)*t^2

t = √( 150/3.5) seconds = 6.54 seconds.

And we know that the initial velocity is:

V0 = (3.5m/s^2)*t

then we can replace the value of t and get:

V0 = (3.5m/s^2)*6.54s = 22.89 m/s