Explanations and Derivations · Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop.

Four Kinematic Equations

Explanations and Derivations

For an object moving with constant acceleration, the average velocity is equal to the average of the initial velocity and the final velocity

We also know, of course, that

Motion Equation #1 Displacement with Constant Acceleration

…move t to the other side...

Motion Equation #2 Velocity with Constant Acceleration

Another Approach to Equation #2

y = mx + b

velo

city

(m/s

)

time (s)

vf = at + vi vf = vi + ator

Motion Equation #3Displacement with Constant Acceleration

By substituting the expression for vf into our first equation for Dx, we get:

First equation:

Motion Equation #4Final velocity after any displacement

See derivation in text book…

FINAL VELOCITY AFTER ANY DISPLACEMENT

vf2 = vi

2 + 2aDx

Summary

Motion with Constant Acceleration

We now have all the equations we need to solve constant-acceleration problems.

vf2 = vi

2 + 2aDx

A car slows down uniformly from a speed of 21.0 m/s to rest in 6.00 seconds. How far did it travel in that time?

The words “slowing down uniformly” implies that the car has a constant acceleration.

vi = +21.0 m/s

t = 6.00 sec

vf = 0 m/s

Dx = ?

Choose an equation

Dx = 63.0 m

A world-class sprinter can burst out of the blocks and reach essentially top speed (of about 11.5 m/s) in the first 15.0 m of the race. What is the average acceleration of this sprinter, and how long does it take her to reach that speed?

vf = +11.5 m/s

Dx = 15.0 m

vi = 0 m/s

a = ?

Choose an equation

= 4.41 m/s2

A world-class sprinter can burst out of the blocks and reach essentially top speed (of about 11.5 m/s) in the first 15.0 m of the race. What is the average acceleration of this sprinter, and

vf = +11.5 m/s

Dx = 15.0 m

vi = 0 m/s

a = 4.41 m/s2

Choose an equation

vf – vi = at

= 2.61 sec

t = ?

Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima accelerates at a rate of 8.00 m/s2, then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.)

vf2 = vi

2 + 2aDx

Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima has an acceleration rate of 8.00 m/s2, determine the displacement of the car during the skidding process.

Dx = 56.3 m

vf2 = vi

2 + 2aDx

Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period.

Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period.

Dx = 50.4 m

A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the car (assume uniform acceleration).

A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the car (assume uniform acceleration).

Dx = 28.6 m

A plane has a takeoff speed of 88.3 m/s and requires 1365 m to reach that speed. Determine the acceleration of the plane and the time required to reach this speed.

vf2 = vi

2 + 2aDx

or

a = 2.86 m/s/st = 30. 9 s

With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object be thrown to reach a height of 91.5 m (equivalent to one football field)? Assume negligible air resistance.

vf2 = vi

2 + 2aDx

With what speed in miles/hr (1 m/s = 2.23 mi/hr) must an object be thrown to reach a height of 91.5 m (equivalent to one football field)? Assume negligible air resistance.

vi = 94.6 mi/hr

• Due Mon:

Kinematic Equations Worksheet