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Transcript of Chapter 2
Motion Along a Straight Line
Younes SinaThe motion of a particle along a straight line at aconstant accelerationis called the "Uniformly Accelerated Motion."
a= constantUniformly Accelerated Motion
Velocity is defined as the change in displacement per unit of time.
V=0 km/hV=260 km/h
A value of 9 (km/h)/s means the velocity changes by 9.0 km/h during each second of the motionExample
t= 29 s
V=0 km/sV= 260 km/sV=0 m/sV= 72 m/sAn acceleration of 2.5 m/s2 is read as 2.5 meters per second per second (or 2.5 meters per second squared) and means that the velocity changes by 2.5 m/s during each second of the motion.
Negative Acceleration and Decreasing VelocityA drag racer crosses the finish line, and the driver deploys a parachute and applies the brakes to slow down (above Figure). The driver begins slowing down when t0 =9.0 s and the cars velocity is v0= 28 m/s. When t= 12.0 s, the velocity has been reduced to v= 13 m/s. What is the average acceleration of the dragster?
Example: A car travels a distance of 350 miles in 5.0 hours. Find its average speed.V=x/t=350 mi/5.0hr=70.0 mi/hrExample :A car's speed changes from 15m/s to 25m/s in along a straight road while moving in one way. Find the magnitude of the acceleration of the car during 5 seconds.A=(vf-vi/tf-ti)=(25m/s-15 m/s)/5.0 s=2.0 m/s2
The Equation of Motion of Uniformly Accelerated Motion:x=(1/2) at2+vitwhere x and t are variables, and a and vi, the constantExample :A car traveling along a straight road at 8.0m/s accelerates to a speed of 15.0m/s in 5.0s. Determineits acceleration, the distance it travels during this period,(c) its equation of motion,(d) the interval of validity of this equation, and(e) the distance already traveled att =2.0 s.
v= 8.0 m/sv= 15.0 m/s
x x= (1/2)a t2+vitCalculation of acceleration using final and initial velocities and time:a= (vf-vi)/tCalculation of traveled distance using acceleration, initial velocity and time:Any object that moves along a straight line and at constant acceleration, has an equation of motion in the form:Part (e) x(2.0)= 0.7 (2.0)2 + 8.0 (2.0) = 18.8m 19m.Part (a):a= (15.0 m/s - 8.0 m/s) / 5.0s = 1.4m/s2.Part (b):
x=(1/2)(1.4m/s2) (5.0s)2+ (8.0m/s)(5.0s) = 58m (rounded to 2 sig. figs.)The equation of motion of this car can be found by substituting the constants in the equation.The constants are:vi= 8.0m/s anda= 1.4m/s2. The equation of motion becomes:Part (c)x(t) =(1/2)(1.4)t2+ 8.0torx(t) = 0.7t2+ 8.0tPart (d) This equation is valid for0 t 5.0s only. (The statement of the problem gives us information for a 5.0s period only).Example: A cyclist traveling at 30.0m/s along a straight road comes uniformly to stop in 5.00s. Determine the stopping acceleration, the stopping distance, and the equation of motion.
v= 30.0 m/s
v= 0 m/st=5.00 s
Solution:a= (vf- vi) / t a= ( 0 - 30.0m/s) / 5.00s =-6.00 m/s2.x= (1/2)a t2+vitx = (1/2)(-6.00m/s2)(5.00s)2 + (30.0m/s)(5.00s) x= +75m.x= (1/2) (-6.00m/s2)t2 + (30.0m/s)t x(t) =-3.00t2 + 30.0t. (a relation betweenxandt).Example : The equation of motion of a lady skiing along a steepand straight ramp isx= 3.40t2+ 2.10t wherexis in meters and tin seconds. Determine(a)her distance from the starting point at the end of 1.0s, 2.0s, 3.0s, 4.0s, and 5.0s. periods. (b) Determine the distances she travels during the 1st, 2nd, 3rd, 4th, and 5th seconds.(c)Are the answers in Part (b) equal? If not why? What should be the pace of motion for such distances to be equal?
1.0 s2.0 s3.0 st= 0x= 3.40t2+ 2.10tx(0) = 3.40(0)2 + 2.10(0) = 0.x(1) = 3.40(1)2 + 2.10(1) = 5.50m.x(2) = 3.40(2)2 + 2.10(2) = 17.8m.x(3) = 3.40(3)2 + 2.10(3) = 36.9m.x(4) = 3.40(4)2 + 2.10(4) = 62.8m.x(5) = 3.4(5)2 + 2.1(5) = 95.5m.d1= x(1) - x(0) = 05.5m - 00.0m = 5.5m.d2= x(2) - x(1) = 17.8m - 05.5m = 12.3m.d3= x(3) - x(2) = 36.9m - 17.8m = 19.1m.d4= x(4) - x(3) = 62.8m - 36.9m = 25.9m.d5= x(5) - x(4) = 95.5m - 62.8m = 32.7m.d1is the distance traveled in the 1st second.d2is the distance traveled in the 2nd second, and so on.(c) As can be seen from Part (b), the distances traveled insubsequent seconds become greater and greater. What is this telling us? It is because the motion is an accelerated one. In accelerated motion, unequal distances are traveled in equal time intervals.
x= (1/2)a t2+vitAn Equation With No Apparent Time Elementa = (vf-vi) / t t=(vf-vi) /a x =(1/2)at2+vit
2ax= vf2-2vfvi+vi2+2vivf-2vi2vf2-vi2=2ax (explicitly independent of time).Example : During take off, an airplane travels 960m along a straight runway to reach a speed of 65m/s before its tires leave the ground. If the motion is uniformly accelerated, determine (a) its acceleration, (b) the elapsed time (c) its equation of motion(d) its midway speed.
x=960mV=0 m/sV= 65 m/sa= constantmidwaySolution: (a) vf2- vi2= 2ax;(65m/s)2- (0m/s)2= 2a(960m) 4225 = 1920aa= 2.2 m/s2(b) a = (vf-vi) / tsolving for ( t ): t = (vf-vi) / at = (65m/s - 0m/s) / ( 2.2m/s2) = 30.sec(c)x(t) = (1/2)( 2.2m/s2)t2 + (0)tx(t) = 1.1t2(d)vf2- vi2= 2ax vf2- (0m/s)2= 2( 2.2 m/s2)[(960/2)m] vf2= 2113 vf= 46 m/s.Freely Falling of Objects:(1)g= (vf-vi)/t(2) y= (1/2)g t2 +vit(3)vf2-vi2=2g yExample:A rock is released from a height of 19.6m.Determine(a) the time it spends in air (b) its speed just before striking the ground.
t=?vf=?vi= 0 m/s(1)g= (vf-vi)/t(2) y= (1/2)g t2 +vit(3)vf2-vi2=2g y(+y)Solution:Since the rock is not thrown and is only released from rest, the initial speed,vi=0.If the (+y) axis is taken to be downward, theng =+ 9.8m/s2, and the equation of motion,y = (1/2) gt2+vitbecomes:y =4.9t2.Substituting the given 19.6m for y, we get:19.6 = 4.9t24 =t2; t = 2.0s. (The positive root is acceptable).To find the velocity at which it hits the ground, we may usevf2-vi2=2g yvf2-02=2 (9.8)(19.6)vf= 19.6 m/s. (Again the positive answer is acceptable). Example: A rifle fired straight upward at the ground level and the bullet leaves the barrel at an initial speed of 315 m/s. Determine the highest it reachesthe time to reach the highest pointthe return time to the ground, and its speed just before striking the ground. Neglect air resistance.
vi= 315 m/svf= 0 m/sh
(+y)Example: A motorcycle ride consists of two segments, as shown in the following Figure. During segment 1, the motorcycle starts from rest, has an acceleration of 2.6 m/s2, and has a displacement of 120 m. Immediately after segment 1, the motorcycle enters segment 2 and begins slowing down with an acceleration of -1.5 m/s2 until its velocity is 12 m/s. What is the displacement of the motorcycle during segment 2?
a=2.6 m/s2vi=0 m/sx=120 mvf2-vi2=2axa=-1.5 m/s2v= 12 m/svf1= ? m/svf2-vi2=2axAnswer: x2=160 m
Example: Using the time and position intervals indicated in the drawing, obtain the velocities for each segment of the trip.
Homework : problems 1 through 5 chapter 2