Chapter 2

Motion Along a Straight Line:

The motion of a particle along a straight line at a constant acceleration is called the "Uniformly Accelerated Motion."

In this chapter, we need to first define velocity and acceleration as was done in Chapter 1.

Velocity:  Velocity is defined as the change in displacement per unit of time.  Note that both velocity and displacement are vector quantities.  Speed is defined as the change in distance per unit of time.  Here both speed and distance are scalar quantities.  The average speed v is mathematically shown as:

Δ pronounced "delta" means a change.  For example, Δx means a change in distance and Δt means a change in time.   Δ also means "final - initial.

Δx = x2 - x1     and     Δt = t2 - t1.

Example 1: A car travels a distance of 350 miles in 5.0 hours.  Find its average speed.

Solution:

Acceleration:  Acceleration is defined as the change in velocity per unit of time.  Again, note that both acceleration and velocity are vector quantities.  Speed that is the magnitude of velocity is a scalar quantity.

Note: For the one dimensional motion of a particle (that means motion along a straight line), and for a one-way motion scenario, the directions of velocity and acceleration do not change.  This allows us to use speed instead of the velocity vector itself and the magnitude of the acceleration vector instead of the acceleration vector itself.

This will be the assumption in almost all of the examples and assigned problems in this chapter.

Example 2:  A car's speed changes from 15m/s to 25m/s in 5.0s along a straight road while moving in one way.  Find the magnitude of the acceleration of the car during this period.

Solution:

Definition of Motion: When the relative position of a particle with respect to a reference point changes with time, the object is said to be in motion.  For convenience, the reference point may be chosen to be the origin, O, in a set of x-y coordinates system.

Equation of Motion:  The equation of motion is defined as the relation between space and time, or simply,  x and t.

The Equation of Motion of Uniformly Accelerated Motion:

For motion along a straight line and at constant acceleration, a, the equation of motion has the form:

The proof of the above equation will be given and discussed later.  Right now, let's just accept this equation There are 2 terms on the right side of it. Verify that each term must have unit of length because the left side is x and has unit of length. You may choose the SI system, for example, and verify that each term has unit of meter.

Example 3: A car traveling along a straight road at 8.0m/s accelerates to a speed of 15.0m/s in 5.0s.   Determine (a) its acceleration, (b) 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 at t = 2.0 s.

Solution: We have so far learned the following equations:  (Rewrite them, but with horizontal fraction bars).

(1)  a = (vf -vi ) /t   ;   (2)  x = (1/2) a t2  +  vi t.    Use horiz. fraction bars.

(a)  a = (15.0m/s - 8.0 m/s) / 5.0s  =  1.4m/s2.

(b)  x (1/2)(1.4m/s2)(5.0s)2  +  (8.0m/s)(5.0s)  =  58m.

Any object that moves along a straight line and at constant acceleration, has an equation of motion in the form:

x = (1/2) a t2 + vit.   The equation of motion of this car can be found by substituting the constants in the equation.

The constants are:  vi = 8.0m/s  and  a = 1.4m/s2.   The equation of motion becomes:

(c)    x(t) = (1/2)(1.4)t2  +  8.0t    or    x(t) = 0.7t2 + 8.0t.      x(t) meanx is a function of t.

(d)         This equation is valid for  0 ≤ t ≤ 5.0s  only.  The statement of the problem gives us information for a 5.0s period only.

(e)         x(2.0) = 0.7 (2.0)2  +  8.0 (2.0)  =  18.8m  ≈ 19m.

Example 4:  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.  Try to solve this on your own before looking at the solution.  Use horizontal fraction bars and not slashes.

Solution:  a =  (vf - vi ) / t  =  ( 0 - 30.0m/s) / 5.00s  =  -6.00 m/s2.

x = (1/2) a t2 + vi t  = (1/2)(-6.00m/s2)(5.00s)2 + (30.0m/s)(5.00s) = +75m.

x =  (1/2) (-6.00m/s2)t2  + (30.0m/s)t     or

x(t) =  -3.00t2  + 30.0t.  (a relation between x and t)

Example 5:  The equation of motion of a lady skiing along a steep and straight ramp is  x = 3.40t2 + 2.10 where x is in meters and t in 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?

Solution:

 (a) x(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. (b) d1 is the distance traveled in the 1st second. d2 is the distance traveled in the 2nd second, and so on. 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.

(c) As can be seen from Part (b), the distances traveled in subsequent seconds become greater and greater.  What is this telling us?  Is it because the motion is an accelerated one?  Ans.:..... .  If your answer is "Yes", you are right. In accelerated motion, unequal distances are traveled in equal time intervals.

An Equation With No Apparent Time Element

Sometimes, in solving problems, we may run into situations where time ( t ) is not given.  It is possible to combine the equations of distance and acceleration to find an equation that is explicitly independent of time.  Look at the following two equations that we already know:  a = (vf -vi) / t   and  x = (1/2)at2 + vit.   If ( t ) is solved for from the 1st equation and substituted into the 2nd equation, the following equation will be resulted:

vf2  -  vi2 = 2ax.    (explicitly independent of time)

Students are encouraged to try the algebra and derive this equation as explained above.

Example 6:  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, and (d) its midway speed.

Solution: (a) vf2 - vi2 = 2ax ;  (65m/s)2 - (0m/s)2 = 2a(960m) ;

4225 = 1920a    ;     a =  2.2 m/s2.

(b)  a = (vf -vi) / ;  solving for ( t )  ;   t = (vf -vi) /a  ;

t = (65m/s - 0m/s) / ( 2.2m/s2)  =  30. sec.

(c)  x(t) = (1/2)( 2.2m/s2) t2  + (0) t     ;    x(t) =  1.1 t2.

(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:

All objects in the vicinity of the Earth are under a gravitational pull toward the center of the Earth whether falling or not.  This gravitational pull imposes an acceleration of  9.8m/s2 on all objects nearby the Earth and the acceleration vector is always toward the Earth's center.  Since falling objects move along straight lines and accelerate as they move, the following equations apply:  Note that these equations are valid only if air resistance can be neglected.  This is true for objects that are tiny, solid, and dense.

(1)   g = (vf -vi ) /t   ;  (2)  y = (1/2) g t2  +  vi t.   Use horiz. fraction bars.

(3)   vf2  -  vi2 = 2 g y.

You already know these equations.  "g", the acceleration of gravity, has replaced "a", and y has replaced x.

 Example 7:  A rock is released from a height of 19.6m.  Determine (a) the time it spends in air and (b) its speed just before striking the ground.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, then g = + 9.8m/s2, and the equation of motion,    y = (1/2) g t2 + vi t , becomes:   y = 4.9t2 . Substituting the given 19.6m for y, we get:   19.6 = 4.9t2  ;  4 = t2 ;  t = 2.0s.  (The positive root is acceptable). To find the velocity at which it hits the ground, we may use  vf2 - vi2 = 2 g y   ;   vf2 - 02 = 2 (9.8)(19.6)  ;  vf = 19.6 m/s. (Again the positive answer is acceptable). It is better to take the origin to be the point of release, or the starting point.  Also, if downward is taken to be the (+ y) axis, then g that is in the same direction of +y becomes positive ( g = +9.8 m/s2).  On this basis, since the rock moves downward it has a positive velocity at any given instant.  Its direction of motion always agrees with the (+ y) axis.

In the following example, it is easy to show that the climbing time and the returning time for an object thrown straight upward are the same.  Also, we will show that the upward throw speed and the downward ground striking speed (on the return) are the same.  Note:  This time, let's assume +y to be upward as usual.  This way, since g always acts downward, it must be given a (-) sign compared to the upward +y.  Draw a figure for the upward motion and a separate figure for the downward motion.

Example 8:  A rifle is fired straight upward at the ground level and the bullet leaves the barrel at an initial speed of 315m/s.   Determine (a) the highest elevation it eaches, (b) the time to reach the highest elevation, (c) the return time to the ground, and (d) its speed just before striking the ground.  Neglect air resistance.

Solution:

(a) At the highest point, the bullet comes to a momentary stop and Vf = 0

Since, we do not know the climbing time, we better use the equation that

does not have time explicitly in it.

Vf2 - Vi2 = 2gy   ;   02 - 3152 = 2 (- 9.8) y    ;   y = 5100m.

(b)  To calculate the climbing time, lets use the equation:

g = (Vf - Vi) /t and solve for t by swapping the diagonal terms.

Make sure to always use horizontal fraction bars.

t = (Vf - Vi) /g     ;        t = (0 - 315) / -9.8  = 32s.

(c)  It is easy to see that the return time is the same as the climbing time of 32s.  When returning, it starts at an initial speed of Vi = 0 and a height of y = 5100m.  Using the y-equation,
t may be calculated as shown:

y = (1/2)g
t2 + Vi t   ;   y = (1/2)g t2  ;  -5100m = (1/2)(-9.8)t2  ;  t = 32s.

(d)  It is easy to show that it strikes the ground at the same speed that it left the rifle:

g = (Vf - Vi) / t   ;  cross-multiplying, we get:   Vf = Vi + g
t.

At the highest point Vi = 0; therefore, Vf = gt.

This gives Vf = (-9.8)(32) = -315m/s.   The (-) sign indicates that the velocity vector is downward.

Chapter 2 Test Yourself 1

1) Motion is (a) the velocity change of an object, (b) the acceleration change of an object, (c) the relative change of the position of an object with respect to a reference point.  To check your answer click here.

2) Equation of motion is a relation between the (a)  constants  (b) the variables distance and time  (c) space and time  (d) both b and c.

3) The equation for acceleration is (a)  a = vf - vi / t     (b)  a = (vf - vi ) / t     (c) a = ( vf + vi ) / t.

4) The equation for speed is (a)  v = x Δ t   (b)  v = Δx  t    (c)  v = (x + t ) / 2.    click here.

5) Uniformly accelerated motion means that the acceleration is (a) constant   (b) zero   (c) varying uniformly.

6) In constant velocity motion, the acceleration is (a) zero  (b) a nonzero constant  (c) distance over time.

7) An object moving at a constant velocity travels (a) different distances in equal intervals of time  (b)equal distances in equal intervals of time  (c) at a constant nonzero acceleration as well.    click here.

8) Change in velocity is (a) Vf + Vi     (b) Vf - Vi      (c) (Vf + Vi ) / 2     (d) Vi - Vf .

9) Average velocity is  (a) Vf + Vi / 2    (b) Vf - Vi     (c) (Vf + Vi ) / 2.

10) A car accelerates uniformly from rest (Vi = 0) to 18m/s in 6.0 seconds.  Its acceleration is (a) 108m/s2  (b)12m/s2  (c)3.0m/s2.

11) A car accelerates uniformly from rest (Vi = 0) to 18m/s in 6.0 seconds.  Its average speed during this period is (a) 9.0m/s  (b) 9.0m/s2  (c) 3.0m/s.    click here.

12) The equation x =  (1/2) a t2  +  vi t  is valid for motion along a straight line if the acceleration is (a) varying  (b) zero  (c) constant   (d) both b and c.

13) Equation x =  (1/2) a t2  +  vi t  simplifies to  x  vi t  if  (a) V is constant  (b) a=0    (c) both a and b.

14) When an object is falling freely under the influence of gravity with negligible air resistance, (a) the acceleration remains constant near the Earth's surface, (b) the acceleration is 9.8 m/s2 downward  (c) the force of gravity on it remains constant because the acceleration is constant,  (d) all of the above.    click here.

15) For straight line motion in one way and at a constant acceleration, the way to calculate an exact average speed is (a) (Vf + Vi ) / 2    (b) Δx / Δt    (c) both a and b.

16) For a uniformly accelerated motion, the object travels (a) equal distances in equal intervals of time  (b) different distances in equal intervals of time  (c) at increased acceleration.

17) An object is released from a height of 490m above the ground.  Assuming Origin to be at the ground level and (+ y) axis to be the usual upward direction,  the equation of motion of the object is  (a) y = 4.9t2   (b) y = 490 - 4.9t2    (c) y = - 4.9t2 + 490t .   click here.

18) An object is released from a height of 122.5m above the ground.  Assuming (+ y) axis to be downward and setting the Origin at where the object is released, the y of the point at which the object strikes the ground is  (a) 0    (b) 122.5m   (c) -122.5m.

19) The time it takes for the object in question 18 to strike the ground is (a)5.0s    (b)10.0s    (c)1.0s.

20) The velocity at which the object in question 18 strikes the ground can be calculated from  the equation (a)  g=(Vf -Vi)/t   (b) vf2-vi2=2gy    (c) both a and b.        click here.

21) When a small rock (negligible air resistance) is thrown straight upward from the ground level, it takes 10.0s to return to the same ground level.  How high does it reach?  (a) 49m    (b) 490m    (c)122.5m.

22) A rock is released from a height of 500m.  The distance it travels in the 3rd second is  (a) 24.5m    (a) 44.1m   (c)167m.

Problems:

1) A driver is to stop his car that is moving at a constant velocity of 24m/s eastward in 4.0 seconds.  If his reaction time is 1.0s, (a) how much time does he have to stop the car and (b) at what minimum acceleration should that happen?   Also, (c) write the equation of motion, and (d) calculate the stopping distance.

2) A train traveling on a straight track at 26.0m/s slows down to 12.0m/s in a distance of 532m.  Find (a) its acceleration, (b) equation of motion, and (c) the elapsed time for this change.  If the point at which slowing down starts is taken to be the origin for this study, find (d) the train distance from the origin at the following instances: 1.0s, 2.0s, 3.0s, 26.0s, 27.0s, and 28.0s.  Name these distance x(1), x(2), x(3), x(26), x(27), and x(28).  (e) Find the distance traveled in the 2nd second, the 26th second,  and the 27th second.

3) A motorcycle is moving along a straight eastward road at a constant speed of 12.0m/s.  Find (a) its acceleration.  (b) if the cyclist keeps this speed for 5.00 minutes, what distance will be traveled during this time?  If the cyclist changes speed to 30.0 m/s in 189m,  find (c) the average speed, (d) the elapsed time, and (e) the acceleration in this phase of motion.

4) Neglecting air resistance, from an initial height of 1.0m above the ground, (a) at what initial speed should a rock be thrown straight upward to come to stop at an elevation of 43m?  (b) Write the equation of motion for the rock assuming that the origin is placed at 1.0m above the ground.  (c) Calculate the ascending time, and (d) find the falling time to the ground as well as the rock's speed just before striking the ground.   Hint: On its way back, place the origin at the new starting point (43.0m elevation) with the (+y) axis upward.  Let g = - 9.81 m/s2.

5) A tiny rock is thrown straight downward from a height of 325m.  If it takes 6.00s to hit the ground, find (a) its initial speed, (b) its final speed just before striking the ground, (c) its equation of motion , and (d) the distance traveled during the last second.  Place the origin at the starting point of the rock with the (+y) axis upward. Neglect air resistance, and let g = -9.81 m/s2.

1) (a) 3.0s    (b) -8.0m/s2    (c) x = - 4.0 t2  +  24 t    (d) 36m

2) (a) -0.500 m/s2    (b) x = - 0.250 t2  +  26.0 t      (c) 28.0s     (d) x(1) = 25.7m;

x(2) = 51.0m;  x(3) = 75.7m;   x(26) = 507m,   x(27) = 520.m,  x(28) = 532m,

(e) 25.3m , 13m, and 12m

3) (a) 0    (b) 3600m    (c) 21m/s    (d) 9.0s    (e) 2.0m/s2

4) (a) 28.7m/s    (b) y = - 4.91 t2  +  28.7 t     (c) 2.93s     (d) 2.96s and 29.0m/s

5) (a) 24.7m/s    (b) 83.6m/s    (c) y = - 4.91 t2  - 24.7 t     (d) 78.8m