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?

Acceleration = -3.5 m/s²

Explanation:

Given:

Initial velocity u = 30 m/s

final velocity velocity v = 23 m/s

Time t = 2 second

Find:

Acceleration

Computation:

v = u + at

23 = 30 + (a)(2)

-7 = 2 a

a = -7 / 2

a = -3.5 m/s²

Acceleration = -3.5 m/s²

resonant frequency

f is prop to m1xm2 double m1 double force