A2500kgCarSlowedDownUniformalyFromAnInitialVelocityOf20.0m/sToTheNorthByA6250NBrakingForceActingOppositeTheCar'sMotion.

Question

What is the car's velocity after 2.50s? How far does the car move during 2.50s? 

Answer 1

Amy, when you have a question, the title should be "Algebra-based physics" or some description of the subject material, and the inside should be the entire question, including the known values. Trying to fit the problem into the title makes it difficult for the tutors to read when we attempt to solve it.

 

You have a two-step problem that involves kinematics and dynamics.

 

First, you calculate the acceleration.

 

F=m*a

 

Substitute your known values in:

 

-6250 N = 2500 kg * a

-2.5 m/s² = a

 

Now that we know a, we can calculate the kinematics.

 

You need V at 2.50 s and d at 2.50 s.

 

V_f = V_i + at

 

Substitute in your known values.

 

V_f = 20.0 m/s + (2.5 s * -2.5 m/s²)

V_f = 20.0 - 6.25

V_f = 13.75 m/s

 

d = V_it + (at²⁾/2

 

d = 20.0 m/s * 2.5 s + (-2.5 m/s² * 2.5 s)/2

d = 50 + -3.125

d = 46.875 m

 

Your velocity after 2.5 seconds is 13.75 m/s and the distance traveled in 2.5 seconds is 46.875 m.