More graphs of motion &

Kinematic equationsSEPTEMBER 16-17, 2014

Review: Slope of a P-T graphWe can calculate the average velocity from a P-T graph by finding the slope of the graph.

Remember: vavg = = = slope of P-T graph

v = 10 m – 0 m

1s – 0 s = 10 m/s

Once we know v, we can plot it on an V-T graph

Review: Slope of a P-T graphWhat if it’s a curvy line?

We can still just pick two points on the graph and use the formula: vavg = =

v = 10 m – 0 m

2s – 1 s = 10 m/s

Once we know v, we can plot it on an V-T graph

Slope of a V-T graphWe can calculate the acceleration from a V-T graph by finding the slope of the graph.

Remember: a = = = slope of V-T graph

a = 10 m/s – 0 m/s

1s – 0 s = 10 m/s2

Once we know a, we can plot it on an A-T graph

What do you think the slope of a V-T graph tells us?

Acceleration!

Area under Motion graphsAnother important thing to know is that the area under the curve in a motion graph has meaning.

area under V-T graph = Δx

Δx = area under V-T graph

How do you find the area of a triangle?

Area = ½ base * height

Δx = ½ * 5 s * 50 m/s = 125 m

Area = Δx

Test your understanding Think about your answer, wait until I tell you, then show me the answer by holding up the correct number of fingers.

.a) Which line has the highest acceleration?

b) How do you know & how could you calculate that acceleration?

c) Which line has the greatest displacement?

d) How do you know & how could you calculate

that displacement?

1

# 1 has the steepest slope. We could calculate a by finding the slope.

# 2 has the greatest displacement.

# 2 has the largest area under the line. We could calculate displacement by finding that area.

You do Use the graph of the motion of a toy train, below, to answer the questions.

1)What is the acceleration between points C and D?

2)How far does the train travel between points A and C?

Kinematic EquationsUp until now, we’ve had just two kinematic equations:

vavg = and a =

These can be re-written in to more user-friendly forms …

Let …

t = the time for which the body acceleratesa = accelerationvi = the velocity at time t = 0, the initial velocityvf = the velocity after time t, the final velocityx = the displacement covered in time tvf = vi + at = x = vi t + 2 vf

2 = vi2 + 2ax

Problem Solving Strategy Diagram the problem List known variables Determine what you are trying to find. Determine your strategy Solve the problem Evaluate your answer. Check whether the units, sign, and

magnitude make sense.

Acceleration Problems – We doGrace is driving her sports car at 30 m/s when a ball rolls out into the street in front of her. Grace slams on the brakes and comes to a stop in 3.0 s. What was the acceleration of Grace’s car?

Sketch:

Known variables:

What are we solving for:

Strategy:

Solution:

Evaluate answer:

Acceleration Problems – We doGrace is driving her sports car at 30 m/s when a ball rolls out into the street in front of her. Grace slams on the brakes and comes to a stop in 3.0 s. What was the acceleration of Grace’s car?

Sketch:

Known variables: Vi = 30m/s Vf = 0 m/s t = 3.0 s

What are we solving for: a

Strategy: Use to solve for a.

Solution: a = (vf – vi)/ t a = (0 m/s-30 m/s) / 3.0 s = -10 m/s2

Evaluate answer: Yes! Units are correct, sign makes sense, and magnitude is reasonable.

vf = vi + at

Acceleration Problems – We doWhat is the displacement after 10.0 s of a mass whose initial velocity is 2.00 m/s and moves with acceleration a = 4.00 m/s2 ?

Sketch:

Known variables:

What are we solving for:

Strategy:

Solution:

Evaluate answer:

Acceleration Problems – We doWhat is the displacement after 10.0 s of a mass whose initial velocity is 2.00 m/s and moves with acceleration a = 4.00 m/s2 ?

Sketch:

Known variables: t = 10.0 s, vi = 2.00 m/s, a = 4.00 m/s2

What are we solving for: x

Strategy:

Solution: 220 m

Evaluate answer: Yes! Distance and sign make sense.

x = vi t + 2

Acceleration Problems – You do1) A car has an initial velocity of 5.0 m/s . When its

displacement increases by 20.0 m, its velocity becomes 7.0 m/s. What is the acceleration?

2) The car accelerate from rest to 28 m/s in 9.0 s. What distance does it travel?

3) A Jet plane lands with a speed of 100 m/s and can accelerate uniformly at a maximum rate of -5.0 m/s2 as it comes to a rest. Can this plane land at an airport where the runway is 0.80 km long?

Acceleration Problems – You do1) A car has an initial velocity of 5.0 m/s . When its displacement

increases by 20.0 m, its velocity becomes 7.0 m/s. What is the acceleration?

Strategy – use Answer: 0.6 m/s2

2) The car accelerates from rest to 28 m/s in 9.0 s. What distance does it travel?

Strategy, find a then find x. Answer: 126 m

3) A Jet plane lands with a speed of 100 m/s and can accelerate uniformly at a maximum rate of -5.0 m/s2 as it comes to a rest. Can this plane land at an airport where the runway is 0.80 km long? Use

Answer: No – it needs 1000 m

vf2 = vi

2 + 2ax

vf2 = vi

2 + 2ax