Question

A car is traveling to the left, which is the negative direction. The direction of travel remains the same throughout this problem. The car's initial speed is 33.6 m/s, and during a 3.42 -second interval, it changes to a final speed of (a) 41.4 m/s and (b) 28.9 m/s. In each case, find the acceleration (magnitude and algebraic sign).

Answer 1

The car is traveling in the negative direction at a speed of 33.6 m/s, so its velocity is -33.6 m/s. We use Vf = Vi + a*t

a) we know we have a:
Vf = -41.4 m/s
Vi = -33.6 m/s
t = 3.42 s

Vf = Vi + a*t
re-arrange for acceleration:
a = (Vf - Vi)/t
a = (-41.4 - (-33.6))/3.42
a = -2.280701754 m/s^2

the car is accelerating because it is changing it is increasing its speed in the same direction (the negative direction)

b)
Vf = -28.9 m/s
Vi = -33.6 m/s
t = 3.42 s

Vf = Vi + a*t
re-arrange for acceleration:
a = (Vf - Vi)/t
a = (-28.9 - (-33.6))/3.42
a = 1.374269006 m/s^2

The car is DE-accelerating because the acceleration is in the opposite direction of the car's motion.