Practice Exam 1

Equations:x = x0 + v0·t + ½a·t2v = v0 + a·tv2 = v0

2 + 2a(x-x0)

v = (v – v0)/2v = ∆x/∆ta = ∆v/∆t

vx = v·cos(θ)vy = v·sin(θ)v2 = vx

2 + vy2

tan(θ) = vy/vx

1) Definitions: define the following terms using words.a) Velocity

b) Acceleration

c) Displacement

d) What is the difference between speed and velocity?

2) Conceptual questions:a) Can an object with a velocity in the positive direction ever have a negative

acceleration?(a) yes(b) no

Explain your reasoning.

(a) A car moving to the right and slowing down will have the accelerationopposite the velocity

b) Two cars emerge side by side from a tunnel. Car A is traveling with aspeed of 60 km/h and has an acceleration of 40 km/h/min. Car B has aspeed of 40 km/h and has an acceleration of 60 km/h/min. Which car ispassing the other as they come out of the tunnel?

(a) Car A(b) Car B

Explain your reasoning(a) They are at the same position, but Car A is traveling faster. Car B wouldcatch up at a latter point.

3) On the left is a qualitative graph of anobject’s velocity. Draw thecorresponding acceleration graph on theright.

4) A ball is rolling across a checked tablecloth. A gray circle shows its position attime equals 0, 1, and 2 seconds. If the grid is 1 cm by 1 cm, calculate thevelocity of the ball. Circle your final answer. (10 points)

v = (x – x0)/t = (13 cm – 3 cm)/2 s = 5 cm/s

5) Describe as precisely as you canhow a cart must move to create thisgraph.

Starting from rest, it speeds up for twoseconds at an acceleration of 2 m/s2. Itthen maintains that speed for 2seconds, then slows down to rest at anacceleration of –2 m/s2.

Velocity

0

+

-

Acceleration

0

-

+

1cm

Position

0

+

-

t=0 s t=1s t=2s

Velocity graph

0

1

2

3

4

5

0 1 2 3 4 5 6

Time (s)

Velo

city

(m/s

)

6) A car traveling 80 km/h hits the brakes and slows down at 4 m/s2. How longdoes it take it to stop? How far did it go in that time?

v0 = 80 km/h = 22.2 m/s x = ?x0 = 0 m t = ?a = -4 m/s2 (slowing down)vf = 0 m/s

vf = v0 + a*t x = x0 + v0*t + ½a*t20 m/s = 22.2 m/s – (4 m/s2)*t x = 0 + 22.2 m/s * 5.55 s + ½(-4 m/s2)*(5.55 s)2

t = 22.2 m/s / 4 m/s2 = 5.55 s x = 61.6 m

7) A mail delivery truck starts out from a stop light and speeds up at a constantrate for until it reaches 18 m/s, the speed it maintains for the rest of 200 meters(until the next major road). If it took 5.0 second to get up to speed, how long didit take to cover the full 200 meters?

Equations to usettotal = t1 + t2 x1 = ½a0*t12 v1 = v0 + a0*t1 x2 = x1 + v1*t2

Solutiont2 = (x2 – x1)/v1t2 = (x2 – ½a0*t12)/v1a0 = v1/t1ttotal = t1 + (x2 – ½( v1/t1)*t12)/v1 = t1 + x2/v1 – ½t1ttotal = 2.5 s + (200 m/18 m/s) = 13.61 s => 14 s

Velocity

0

+

-

Chart of variablest0 = 0.0 s t1 = 5.0 s t2 = ?x0 = 0.0 m x1 = ? x2 = 200 mv0 = 0.0 m/s v1 = 18 m/s v2 = 18 m/sa0 = ? a1 = 0.0 m/s2

ttotal = t1 + t2

8) Real world problem. Fill in ONLY the indicated steps of the FOCUS andDESCRIBE steps. DO NOT SOLVE THE PROBLEM.

Michael Jordan, practicing for a possible comeback, is practicing three pointshots. He is 25 feet from the basket, which is 10 feet above the court. There is adefender directly in front of him, so he first fakes a move with his head and thenleaps straight up and shoots the ball from a point about 8 feet above the court.The ball falls through the basket 1.3 seconds later. What was the velocity of theball in the vertical and horizontal directions as it left his hand?

Natural descriptionSketch with Given Information

PHYSICS DESCRIPTIONDiagram Make a chart of known and unknown

variables. Indicate which one you will solvefor.

Write down the general equations. (Write down ONLY the equations needed to solve this problem.)

x = x0 + vxty = y0 + vy0t + ½at2

8 feet

10 feet

25 feet

Horizontal Verticalx0 = 0 y0 = 8 ftx = 25 ft. y = 10 ftvx = ? vy0 = ?ax = 0 ft/s2 ay = -32 ft/s2

t = 1.3 s

Horizontal position

0

+

Vertical position

0

+