What is the maximum speed at which a car could be moving and not hit a barrier 35.0 m ahead if the average acceleration during braking is -10.0 m/s2 and it takes the driver 0.90 s before he applies the brakes?
subtract from the 35 m the distance during reactiontime (v*.9) so the stopping distance is 35-.9v.
That is equal to d in ...
vf^2=v^2+2ad
vf is zero, solve for v.
35 - (.9)(-10.0) = 44
V = 29.664 and this is incorrect
Do I use the acceleration for the velocity or do I figure out the velocity from the informatin given?
35 - (.9)(-10.0) = 44
V = 29.664 and this is incorrect
Do I use the acceleration for the velocity or do I figure out the velocity from the informatin given?
To answer this question, we need to use the kinematic equation that relates initial velocity, final velocity, acceleration, and displacement:
\(d = v_i * t + \frac{1}{2} * a * t^2\),
where:
\(d\) is the displacement (35.0 m),
\(v_i\) is the initial velocity,
\(a\) is the acceleration (-10.0 m/s\(^2\)),
\(t\) is the time taken by the driver to apply the brakes (0.90 s).
We can rearrange the equation to solve for the initial velocity (\(v_i\)):
\(v_i = \frac{d - \frac{1}{2} * a * t^2}{t}\).
Now, substitute the given values into the equation:
\(v_i = \frac{35.0 - \frac{1}{2} * (-10.0) * (0.90)^2}{0.90}\).
Calculating the equation:
\(v_i = \frac{35.0 - \frac{1}{2} * (-10.0) * 0.81}{0.90}\),
\(v_i = \frac{35.0 + 0.405}{0.90}\),
\(v_i = \frac{35.405}{0.90}\),
\(v_i \approx 39.34 \, m/s\).
Therefore, the maximum speed at which the car can be moving without hitting the barrier 35.0 m ahead is approximately 39.34 m/s.