acceleration kinematics

Question

While driving to work at 25 m/s [E], Jamaal sees a squirrel run out onto the street in front of him chasing an errant nut. Not wanting to hit the squirrel, Jamaal applies the brakes for 2.00 s to slow the car down over the next 37.0 m and allows the squirrel enough time to get out of the way. (5 marks)

a. What is the acceleration of the car?

b. What is the car’s final velocity?

Answer 1

Hi Stephen! To begin this problem, let's summarize what we already know and what we're looking for. The problem provides us with Jamaal's initial velocity and the distance and time for which he slows down. With that information, we want to calculate the acceleration and final velocity of the car.

Part a) To calculate the acceleration of the car as Jamaal brakes, we can use the following equation,

Δx = v₀t + (1/2)at²,

where Δx is the change in displacement, v₀ is the initial velocity, t is time, and a is the acceleration. We are provided with all of these values with the exception of the acceleration. Plugging in what we're given in the problem, we get

37 = 25 * 2 + (1/2) * a * 2².

Then, solving for a, our acceleration, we get

37 = 50 + (1/2) * 4 * a

37 = 50 + 2 * a

-13 = 2 * a

-6.5 = a.

So, we are able to find out the acceleration equals -6.5 m/s².

Part b) When we start part b of this problem, something worth noting is that we now know our acceleration in addition to the rest of the values provided to us in the problem, so we can use that to our advantage when calculating Jamaal's final velocity.

To find this final velocity, we can use the following equation,

Δv = v_f - v_i = at,

where Δv is the change in velocity, v_f is the final velocity, v_i is the initial velocity, a is the acceleration, and t is time. Plugging in what we know so far, we get

v_f - 25 = -6.5 * 2.

Then, solving for our final velocity v_f, we get

v_f - 25 = -13

v_f = 12.

So, we are able to find out that the final velocity equals 12 m/s.