**Acceleration Formula**

Acceleration is a measure of how quickly the velocity of an object changes. So, the acceleration is the change in the velocity, divided by the time. Acceleration has a magnitude (a value) and a direction. The direction of the acceleration does not have to be the same as the direction of the velocity. The units for acceleration are meters per second squared (m/s^{2}).

*a* = acceleration (m/s^{2})

*v _{f}* = the final velocity (m/s)

*v _{i}* = the initial velocity (m/s)

*t* = the time in which the change occurs (s)

Δ*v* = short form for “the change in” velocity (m/s)

**Acceleration Formula Questions:**1) A sports car is travelling at a constant velocity *v = 5.00 m/s*. The driver steps on the gas, and the car accelerates forward. After 10.0 seconds, the driver stops accelerating and maintains a constant velocity *v = 25.0 m/s*. What was the car’s acceleration?

Answer: The initial velocity is *v _{i} = 5.00 m/s*, in the forward direction. The final velocity is

*v*in the forward direction. The time in which this change occurred is 10.0

_{f}= 25.0 m/s*s*. The acceleration is in the forward direction, with a value:

The car’s acceleration is 2.00* m/s ^{2}*, forward.

2) A child drops a rock off of a cliff. The rock falls for *15.0 s* before hitting the ground. The acceleration due to gravity is *g = 9.80 m/s ^{2}*. What was the velocity of the rock the instant before it hit the ground?

Answer: The rock was released from rest, so the initial velocity is *v _{i} = 0.00 m/s*. The time in which the change occurred is

*15.0 s*. The acceleration is

*9.80 m/s*. The final velocity must be found, so rearrange the equation:

^{2}*v _{f} = v_{i} + at*

*v _{f} = 0.00 m/s +(9.80 m/s^{2})(15.0 s)*

*v _{f} = 147 m/s*

The rock is falling, so the direction of the velocity is down.