Question

A car traveling east in a straight line on a highway decreases its speed from 30. meters per second to 23. meters per second in 2.0 seconds. What is the magnitude of the car's average acceleration to the nearest tenths place, during this time interval?

Answer 1

-3.5m/s²

Step-by-step explanation:

Given

Initial Velocity, u = 30m/s

Final Velocity, v = 23m/s

Time, t = 2.0s

Required

Determine the magnitude of acceleration

This is calculated using first equation of motion.

v = u + at

Substitute values for v, u and t

23 = 30 + a * 2.0

23 = 30 + 2.0a

23 - 30 = 2.0a

-7 =2.0a

Solve for a

a = -7 ÷ 2.0

a = -3.5m/s²