Motion In One Dimension

Physics Unit 1 Topic 2 Motion in One Dimension

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1. Displacement The displacement is a vector that points from an object’s initial position to its final position and has a magnitude that equals the shortest distance between the two positions.

2. Speed and Velocity AVERAGE SPEED The distance traveled divided by the time required to cover the distance. Average speed = ___Distance ___ Elapsed time Example 1 How far does a jogger run in 1.5 hrs (5400 s) if his average speed is 2.22 m/s? Solution:

Distance = Average speed x Elapsed time = 2.22 m/s x 5400 s = 11,988 m Answer 2

AVERAGE VELOCITY Average velocity is displacement (∆x) divided by elapsed time (∆t). Average velocity = _Displacement__ Elapsed time

v = _ x – x0_ = _ ∆x__ t – t0

∆t

Example 2. A driver runs a jet-engine car through a course in two runs, one in each direction, to nullify wind effects. The car first travels from left to right and covers a distance of 1,609 meters in a time of 4.740 sec. In the reverse direction, the car travels the same distance in 4.695 s. Determine the average velocity for each run. Solution: Run 1

v = ∆x/∆t = 1,609 m / 4.740 s = 339.5 m/s

Run 2

v = ∆x /∆t = -1,609 m / 4.695 s = -342.7 m/s

Note: The magnitudes of the velocities are 339.5 and 342.7 m/s. The average is 341.1 m/s

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Solve: 3.A fishing boat leaves port at 04:30 AM in search of the day’s catch. The boat travels 4.5 km [E], then 2.5 km [S] and finally 1.5 km [W] before discovering a large school of fish on the sonar screen at 06:30 AM. (a) Draw a vector scale diagram of the boat’s motion. (b) Calculate the boat’s average speed. (c) Determine the boat’s average velocity.+ 2. A straight track is 1600 m in length. A runner begins at the starting line, runs due east for the full length of the track, turns around, and runs halfway back. The time for this run is five minutes. What is the runner’s average velocity, and what is his average speed? 3. A woman and her dog are out for a morning run for the river 4.0 km away. The woman runs at 2.5 m/s in a straight line. The dog is unleashed and runs back and forth at 4.5 m/s between his owner and the river, until she reaches the river. What is the total distance run by the dog?

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INSTANTANEOUS VELOCITY Instantaneous Velocity

indicates how fast a particle moves and the direction of the motion at each instant of time;

Instantaneous Speed

the magnitude of the instantaneous velocity. In a moving car, it is the number (with units) indicated by the speedometer.

The instantaneous velocity is approximately equal to (≈) average velocity for sufficiently small ∆t, and, in the limit that ∆t becomes infinitesimally small, the instantaneous velocity and the average velocity become equal, i.e.,

v = lim (∆x /∆t) ∆t

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3. Acceleration Acceleration

the rate at which the velocity of a particle changes with time.

Average Acceleration (ā)

change of velocity divided by the corresponding time interval, i.e.,

ā = _v2 – v1_ = _∆v_ t2 – t1 = ∆t

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Example 5. A drag racer crosses the finish line, and the driver deploys a parachute and applies the brakes to slow down. The driver begins slowing down when t0 = 9.0 s and the car’s velocity is v0 = 28 m/s, When t = 12 s, the velocity has been reduced to v = + 13 m/s. What is the average acceleration of the dragster? Given: Solution:

t0 = 9.0 s; v0 = 28 m/s; v = +13 m/s ; t =

ā = _v – v0_ = _13 – (+28)_ = - 15/3 m/s2 t – t0

12 – 9

= - 5 m/s2 CYU 2: Two cars are moving on straight sections of a highway. The acceleration of the first car is greater than the second car , and both accelerations have the same direction. Which one of the following is true? (a) the velocity of the first car is always greater than the velocity of the second car. (b) the velocity of the second car is always greater than the velocity of the first car. (c) In the same interval, the velocity of the first car changes by a greater amount than the velocity of the second car does. (d) In the same interval, the velocity of the second car changes by a greater amount than the velocity of the first car does. [c]

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4. Equations of Kinematics for Constant Acceleration For an object with initial velocity of v0 at t0 = 0 and moves for a time t with a constant acceleration a, the final velocity is found by: ā = a = v – v0 _ or v = v0 + at t

(constant acceleration)

(2.4)

The displacement x at time t, assuming that x0 = 0 m at t0 = 0 s, is: vave = x – x0_ = x_ or x = vavet t –t0 t Because the acceleration is constant, the average velocity increases at a constant rate. Thus, the average velicity is midway between the initial and final velocities: vave = ½(v0 + v)

(constant acceleration)

The displacement at time t is now determined as: x = vavet = ½(v0 + v)t

(constant acceleration)

(2.7)

Note that there are five kinematic variables in equations (2.4) and (2.7): 1. x = displacement

4. v0 = initial velocity at t0 = 0

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Each of equations (2.4) and (2.7) contains four of these variables, so if three of them are known, the fourth can always be found. Example: 1. A speed boat has a constant acceleration of +2.0 m/s2 . If the initial velocity of the boat is +6.0 m/s, find its displacement after 8 seconds. Solution: Given:

a = +2.0 m/s2 ; v0 = +6.0 m/s; t = 8 s. From eq. (2.7),

x = ½ (v0 + v)t

from eq. (2.4):

v = v0 + at

v = 6.0 m/s + 2.0 m/s2 (8 s) = 22 m/s x = ½(6.0 m/s + 22 m/s) 8 s = 112 m ≈ 110 m (correct to 2 dec places) Note: Equation (2.4) may be substituted into equation (2.7) to yield an equation that allows solving for x in a single step: x = ½[v0 + (v0 + at)]t = ½(2v0t + at2) x = v0t + ½at2

(2.8) 9

2. A jet is taking off the deck of an aircraft carrier. Starting from rest, the jet is catapulted with a constant acceleration of +31 m/s2 along a straight line and reaches a velocity of 62 m/s. Find the displacement of the jet. Given:

v0 = 0; a = +31 m/s2; v = 62 m/s.

x=?

Solution: x = v0t + ½ at2

(1)

from equation (a): v = v0 + at; t = (v – v0)_ = 62 m/s – 0_ a 31 m/s2 t = 2 sec Subst in (1) :

x = ½ ( 31 m/s2)(2 s)2 = 62 m Ans

Notice that solving Equation 2.4 for the time [ t = (v – v0.)/a] and then substituting into Equation 2.7, and solving for v2 yields: v2 = v02 + 2ax

(2.9)

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Equations of Kinematics for Constant Acceleration No.

Equation

2.4

Variables x

a

v

v0

t

v = v0 + at

-

√

√

√

√

2.7

x = ½ (v0 + v)t

√

-

√

√

√

2.8

x = v0 t + ½ at2

√

√

-

√

√

2.9

V2 = v02 + 2ax

√

√

√

√

-

Deceleration Versus Negative Acceleration Deceleration:

means only that the acceleration vector points opposite to the velocity vector and indicates that the moving object is slowing down. When a moving object slows down, its instantaneous speed decreases. A decelerating object does not necessarily have a negative acceleration.

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When a moving object slows down, its instantaneous speed (the magnitude of the instantaneous velocity) decreases. One possibility is that the velocity vector of the object points to the right, in the positive direction. The term “decelerating” implies that the acceleration vector points opposite, or to the left, which is the negative direction. Here, the value of the acceleration would indeed be negative. However, the object could be travelling to the left, in which case the acceleration vector would point opposite or to the right, so the acceleration a would have a positive value. Example. A motorcycle, starting from rest, has an acceleration of +2.6m/s2 . After the motorcycle travelled a distance of 120 m, it slows down with an acceleration of -1.5 m/s2 until its velocity is +12m/s. What is the total displacement of the motorcycle? Solution: Phase 1:

v2 = v02 + 2ax ; v2 = 0 + 2(2.6m/s2)(120m) = 624 (m/s)2 v = 25 m/s this is the initial velocity for phase 2.

Phase 2:

x = (v02 – v2)/2a = [(12)2 + (25)2)]/2(-1.5 m/s2) = 160 m Total displacement = 120 + 160 = 280 m 12