university physics: mechanics

University Physics: Mechanics

Ch2. STRAIGHT LINE MOTION

Lecture 2

Dr.-Ing. Erwin Sitompulhttp://zitompul.wordpress.com

2014

2/2Erwin Sitompul University Physics: Mechanics

Solution for Homework 1: Truck

You drives a truck along a straight road for 8.4 km at 70 km/h, at which point the truck runs out of gasoline and stops. Over the next 30 min, you walk another 2.0 km farther along the road to a gasoline station.

(a) What is your overall displacement from the beginning of your drive to your arrival at the station?

2 8.4 2.0 10.4 kmx 1 0x

2 1 10.4 km 0 10.4 kmx x x

(b) What is the time interval Δt from the beginning of your drive to your arrival at the station?

drv wlkt t t drv

8.4 km

70 km/ht 0.12 h

wlk 30 mint 0.5 h 0.12 h 0.5 h 0.62 h



(c) What is your average velocity vavg from the beginning of your drive to your arrival at the station? Find it both numerically and graphically.

avg

xv

t

10.4 km

0.62 h

16.77 km/h


(d) Suppose that to pump the gasoline, pay for it, and walk back to the truck takes you another 45 min. What is your average speed from the beginning of your drive to you return to the truck with the gas?


avg

s

t

total distance

8.4 2 2 12.4 km total distance

0.12 0.5 0.75 1.37 hr total time interval

12.4 km

1.37 hr 9.05 km/h


Instantaneous Velocity and Speed Instantaneous velocity (or

simply velocity) is the average velocity over a very short period of time interval

0limt

xv

t

dx

dt

Velocity v at any instant is the slope of the position-time curve.

Instantaneous speed (or simply speed) is the magnitude of velocity, that is, speed is velocity without any indication of direction

dx

dtspeed

● What is the car’s vavg?


instantaneous velocity v = slope of curve at any point

Instantaneous Velocity


Example: Elevator Cab

Figure (a) on the left is an x(t) plot for an elevator cab that is initially stationary, then moves upward (which we take to be the positive direction of x), and then stops. Plot v(t).

xv

t

The slope of x(t), which is the v(t), is zero in the intervals from 0 to 1 s and from 9 s on the cab is stationary in these intervals.

During the interval bc, the slope is nonzero constant the cab moves with constant velocity.

24 44m s

8 3

Between 1 s and 3 s the cab begins to

move, and between 8 s and 9 s it slows down the velocity of the cab varies. ● Given v(t), can you

determine x(t) exactly?


Motion with constant velocity on x-t graph

Motion with Constant Velocity

dxv

dt constant

dx v dt

x vt c

Taking at time t0 = 0 the position is at x0 c = x0

dx v dt

0x x vt


Position, Time, and Velocity

What is the velocity a cyclist for each stage of this trip?


Acceleration Average acceleration is

the ratio of change in velocity to the time interval.

2 1avg

2 1

v va

t t

Instantaneous acceleration (or simply acceleration) is the derivative of the velocity with respect to time.

0limt

va

t

dv

dt

v

t

dva

dt

d dxdt dt

2

2

d x

dt


Acceleration

Compare the a(t) curve with the v(t) curve.Each point on the a(t) curve shows the

derivative (slope) of the v(t) curve at the corresponding time.

When v is constant, the derivative is zero a is also zero.

When the cab first begin to move, the v(t) curve has a positive derivation, which means that a(t) is positive.

When the cab slows to a stop, the derivative of v(t) is negative; that is, a(t) is negative.


Motions with constant acceleration on v-t graph

Motion with Constant Acceleration

dva

dt constant

dv a dt

dv a dt v at c

Taking at time t0 = 0 the velocity equals v0 c = v0

0v v at


Motions with constant acceleration on v-t graph


0v v at

dx v dt

0( ) dx v at dt

0( ) dx v at dt 2

0

1

2x v t at c

Taking at time t0 = 0 the position is at x0 c = x0

20 0

1

2x x v t at


a constant

0v v at

20 0

1

2x x v t at

From these two equations, the following equations can be derived:

2 20 02 ( )v v a x x

0 0

1( )

2x x v v t

20

1

2x x vt at



Example: Particle’s MovementThe position of a particle is given by x = 4t2 – 2t + 10, where x is the distance from origin in meters and t the time in seconds.

.

(a) Find the displacement of the particle for the time interval from t = 1 s to t = 2 s.

(b) Find also the average velocity for the above given time interval.

(c) Find the instantaneous velocity of the particle at t = 0.5 s.

21 4(1) 2(1) 10 12x

22 4(2) 2(2) 10 22x

2 1 22 12 10 mx x x

2 1avg

2 1

x xv

t t

22 12

10 m s2 1

dxv

dt 8 2t

8(0.5) 2 2m sv At t = 0.5 s,


What is velocity in intervals A, B, C, D

What is acceleration in intervals A, B, C

2 m/s

Questions

0 m/s 1 m/s

–1.5 m/s 2 m/s2 0 m/s2

–0.5 m/s2


Example: Porsche

Spotting a police car, you brake a Porsche from a speed of 100 km/h to a speed of 80.0 km/h during a displacement of 88.0 m, at a constant acceleration.

(a) What is that acceleration?2 2

0 02 ( )v v a x x 2 222.22 27.78 2 (88 0)a

2 222.22 27.78

(2)(88)a

21.58 m/s

1000 m80 km/h 80

3600 s

1000 m100 km/h 100

3600 s

27.78 m/s

22.22 m/s

2 20 0 0

1 120 0 0 02 2

1 20 2

2 ( )

( )

v v at v v a x x

x x v t at x x v v t

x x vt at


Spotting a police car, you brake a Porsche from a speed of 100 km/h to a speed of 80.0 km/h during a displacement of 88.0 m, at a constant acceleration.

(b) How much time is required for the given decrease in speed?

0v v at

0v vt

a

22.22 27.78

1.58

3.52 s

2 20 0 0

1 120 0 0 02 2

1 20 2

2 ( )

( )

v v at v v a x x


x x vt at

Example: Porsche


Car accelerating and decelerating

Example: Porsche


a > 0, a = 0, a < 0


Example: Ferrari

A Ferrari F430 needs only 3.50 s to accelerate from 0-100 km/h.

What is its average acceleration?

What is the street distance required by the car to do such acceleration?

2 20 0 0

1 120 0 0 02 2

1 20 2

2 ( )

( )

v v at v v a x x


x x vt at


Velocity vs. Time Position vs. Time

Both cars move with constant velocity

Red car starts moving 4 seconds after the blue car

Both car meet at t = 7 s.

Illustration: Overtaking Maneuver

● Can you determine the exact value of vblue car?


Solution:Simple. Just ask the merry-go-round operator to stop!

Trivia: How to Stop?

You are driving a car with ever changing velocity but constant speed of 2 m/s. On your right is a steep cliff with the height of 50 cm. Directly in front of you there is a horse and behind you an elephant, both of which travel at your own speed. On your left there is a fire truck blocking you. How do you stop your car?


Which time periods indicates that an object moves at constant speed?

What are (a) Initial direction of travel?(b) Final direction of travel?(c) Does the particle stop

momentarily?(d) Is the acceleration positive

or negative?(e) Is the acceleration

constant or varying?

Questions

E, where a = 0

– +

Yes

+

Constant


How far does the runner travel in 16 s?

Questions

10 02

102

( )

( )

x x v v t

x v v t

11 2

(0 8) 28

x

12 2

(8 8) 864

x

13 2

(8 4)212

x

14 2

(4 4)416

x

total 1 2 3 4x x x x x


Example: RaceA caravan moves with a constant velocity of 60 km/h along a straight road when it passes a roadster which is at rest. Exactly when the caravan passes the roadster, the roadster starts to move with an acceleration of 4 m/s2.

(a) How much time does the roadster need to catch up the caravan?

0

01 2

0 0 22 2

0 01

0 021 2

0 2

2 ( )

( )

x x vt

v v at

x x v t at

v v a x x

x x v v t

x x vt at

caravan roadsterx x

0 00, 0x t

1 20,caravan caravan 0,roadster 0,roadster roadster2x v t x v t a t

1 2

20 16.67 0 0 4t t t 22 16.67 0t t ( 8.33) 0t t 1 20, 8.33 st t

caravan 60km h 16.67 m sv

Thus, the roadster will catch up the caravan after 8.33 s.


Example: RaceA caravan moves with a constant velocity of 60 km/h along a straight road when it passes a roadster which is at rest. Exactly when the caravan passes the roadster, the roadster starts to move with an acceleration of 4 m/s2.

(b) How far does the roadster already move when it catches up the caravan?

0

01 2

0 0 22 2

0 01

0 021 2

0 2

2 ( )

( )

x x vt

v v at

x x v t at

v v a x x

x x v v t

x x vt at

caravan 0,caravan caravanx x v t

0 0x

Both the vehicles travel 138.9 m before they pass each other again.

Distance traveled

0 (16.67)(8.33)

1 2roadster 0,roadster 0,roadster roadster2x x v t a t

1 2

20 (0)(8.33) (4)(8.33) 138.9 m

138.9 m


Homework 2: Aprilia vs. KawasakiAn Aprilia and a Kawasaki are separated by 200 m when they start to move towards each other at t = 0.

(a) Determine the point where the two motorcycles meet each other.

200 m

The Aprilia moves with initial velocity 5 m/s and acceleration 4 m/s2. The Kawasaki runs with initial velocity 10 m/s and acceleration 6 m/s2.

(b) Determine the velocity of Aprilia and Kawasaki by the time they meet each other.


Homework 2A: Running ExerciseYou come late to a running exercise and your friends already run 200 m with constant speed of 4 m/s.

The athletic trainer orders you to catch up your friends within 1 minute.

(a) If you run with constant speed, determine the minimum speed you have to take so that you can fulfill the trainer’s order.

(b) If you run with minimum speed, determine the point where you catch up your friends.


Homework 2B: Runway Length1. An electron moving along the x axis has a position given by

x = 16te–t m, where t is in seconds. How far is the electron from the origin when it momentarily stops?

2. An airplane lands at a speed of 160 mi/h and decelerates at the rate of 10 mi/h/s. If the plane travels at a constant speed of 160 mi/h for 1.0 s after landing before applying the brakes, what is the total displacement of the aircraft between touchdown on the runway and coming to rest?

university physics: mechanics

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