university physics: mechanics

University Physics: Mechanics

Ch2. STRAIGHT LINE MOTIONLecture 2

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

2/2Erwin Sitompul University Physics: Mechanics

Solution for Homework 1: TruckYou 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 drv8.4 km

70 km/ht 0.12 h

wlk 30 mint 0.5 h 0.12 h 0.5 h 0.62 h


Solution for Homework 1: Truck(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.

avgxvt

10.4 km0.62 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?

Solution for Homework 1: Truck

avg st

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 km1.37 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

xvt

dxdt

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

dxdt

speed


Motion with constant velocity on x-t graph

Motion with Constant Velocity

dxvdt

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


Acceleration Average acceleration is

the ratio of change in velocity to the time interval.

2 1avg

2 1

v vat t

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

0limt

vat

dvdt

vt

dvadt

d dxdt dt

2

2

d xdt


Motions with constant acceleration on v-t graph

Motion with Constant Acceleration dvadt

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

Motion with Constant Acceleration

0v v at

dx v dt

0( ) dx v at dt

0( ) dx v at dt 2

012

x v t at c

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

20 0

12

x x v t at


Motion with Constant Acceleration

a constant

0v v at

20 0

12

x x v t at

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

2 20 02 ( )v v a x x

0 01 ( )2

x x v v t

20

12

x x vt at


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: PorscheSpotting 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 27.782 88

a

21.58 m/s

1000 m80 km/h 803600 s

1000 m100 km/h 1003600 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 xx x v t at x x v v t

x x vt at


Example: PorscheSpotting 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 vta

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 xx x v t at x x v v t

x x vt at


Example: Porsche

Car accelerating and decelerating


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) 2

8x

12 2 (8 8) 8

64x

1

3 2 (8 4)212

x

14 2 (4 4)4

16x

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 vtv v at

x x v t atv v a x xx x v v tx 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

1 220 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 vtv v at

x x v t atv v a x xx x v v tx 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

1 220 (0)(8.33) (4)(8.33)

138.9 m

138.9 m


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 xvt t

22 12 10 m s

2 1

dxvdt

8 2t

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


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 2

You 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) 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.

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