A car is traveling at 26.0 m/s when the driver suddenly applies the brakes, giving the car a constant deceleration. The car comes to a stop in a distance of 120.0 m. How fast was the car moving when it was 30.0 m past the point where the brakes were applied?
A) 7.50 m/s
B) 15.0 m/s
C) 23.5 m/s
D) 22.5 m/s
E) 28.0 m/s

Question

Quizplus · Accepted Answer

First, use the equation of motion: 
$v^2 = u^2 + 2as$
where
v = final velocity (0 m/s)
u = initial velocity (unknown)
a = acceleration (constant deceleration in this case)
s = distance (120.0 m)

Rearranging the equation, we have: 
$u = \sqrt{v^2 - 2as}$

Substituting the given values, we get:

$u = \sqrt{(26.0)^2 - 2(-a)(120.0)}$

We don't know the value of a, but we can use another equation that involves distance and acceleration:

$s = ut + \frac{1}{2}at^2$

At the point where the brakes were applied, the car had traveled 30.0 m further. Therefore, the distance traveled before coming to a stop is:

$s = 120.0 - 30.0 = 90.0$

Substituting this value and rearranging the equation, we get:

$a = \frac{2(s - ut)}{t^2}$

We don't know t, but we can eliminate it by using another equation that relates initial and final velocities, acceleration, and distance:

$v^2 = u^2 + 2as$

Rearranging the equation, we get:

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

Substituting this expression for t into the equation for a, we get:

$a = \frac{2(s - ut)}{(v - u)^2}$

Now we have two equations with two unknowns (u and a). Solving simultaneously gives:

$u = 22.5$ m/s and $a = 4.02$ m/s$^2$

Therefore, the answer is D) 22.5 m/s.

A Car Is Traveling at 26