mechanics unit 5: motion and forces 5.6 motion in one dimension - speed and velocity, acceleration

Mechanics

Unit 5: Motion and Forces

5.6 Motion in one Dimension - Speed and Velocity, Acceleration

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Fundamentals of physics Motion in One Dimension

In this chapter we discuss motion in one dimension. We introduce definitions for displacement, velocity and acceleration, and derive equations of motion for bodies moving in one dimension with constant acceleration. We apply these equations to the situation of a body moving under the influence of gravity alone.

----------------------------------------------------- Displacement Average Velocity Instantaneous Velocity Acceleration One Dimensional Motion with Constant Acceleration

Derivation of Kinematic Equations of Motion Freely Falling Bodies

Problems

Fundamentals of physics Motion in One Dimension

-----------------------------------------------------Definition: Displacement is change in position,

where is the final position and is the initial position. The arrow indicates that displacement is a vector quantity: it has direction and magnitude. In ONE dimension, there are only two possible directions which can be specified with either a plus or a minus sign. Other examples of vectors are velocity, acceleration and force. In contrast, scalar quantities have only magnitude. Some examples of scalars are speed, mass, temperature and energy.

Note: the overbar is frequently used to denote an average quantity t is always > 0 so the sign of depends only on the sign of x .

Fundamentals of physics Average Velocity

-----------------------------------------------------Definition: Average Velocity is displacement over total time. Mathematically:

Eq. 2.1

Fundamentals of physics Average Velocity

-----------------------------------------------------Graphical interpretation of velocity: Consider 1-d motion from point P (with coordinates xi, ti) to point Q (at xf, tf). We can plot the trajectory on a graph (see Figure 2.1).

Then from the previous equation is just the slope of the line joining P and Q.

Figure 2.1: Graphical interpretation of velocity

Fundamentals of physics Instantaneous Velocity

-----------------------------------------------------Definition: Instantaneous velocity is defined mathematically:

Example: Table 2.1 gives data on the position of a runner on a track at various times.

Find the runner's instantaneous velocity at t = 1.00 s. As a first estimate, find the average velocity for the total observed part of the run. We have,

Table 2.1: Position and time for a runner.

Eq. 2.3

Eq. 2.2


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From the definition of instantaneous velocity Eq.(2.2), we can get a better approximation by taking a shorter time interval. The best approximation we can get from this data gives,

Table 2.1: Position and time for a runner.

Eq. 2.4


-----------------------------------------------------We can interpret the instantaneous velocity graphically as follows. Recall that the average velocity is the slope of the line joining P and Q (from Figure 2.1). To get the instantaneous velocity we need to take t0, or PQ. When PQ, the line joining P and Q approaches the tangent to the curve at P (or Q). Thus the slope of the tangent at P is the instantaneous velocity at P. Note that if the trajectory were a straight line, we would get v = , the same for all t . Note: Instantaneous velocity gives more information than average velocity. The magnitude of the velocity (either average or instantaneous) is referred to as the speed.

Fundamentals of physics Acceleration

-----------------------------------------------------Definition: Average acceleration is the change in velocity over the change in time:

Definition: Instantaneous acceleration is calculated by taking shorter and shorter time intervals, i.e. taking t0:

Eq. 2.5

Eq. 2.6

Fundamentals of physics Acceleration

-----------------------------------------------------Note: Acceleration is the rate of change of velocity. When velocity and acceleration are in the same direction, speed increases with time. When velocity and acceleration are in opposite directions, speed decreases with time. Graphical interpretation of acceleration: On a graph of v versus t , the average acceleration between P and Q is the slope of the line between P and Q, and the instantaneous acceleration at P is the tangent to the curve at P.

From now on ``velocity'' and ``acceleration'' will refer to the instantaneous quantities.

Fundamentals of physics One Dimensional Motion with Constant Acceleration

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Constant acceleration means velocity increases or decreases at the same rate throughout the motion.

Example: an object falling near the earth's surface (neglecting air resistance).

Fundamentals of physics Derivation of Kinematic Equations of Motion

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Equation of motion ( Eq2.7)

So we get


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So

Now

and

Which gives Equation of motion ( Eq2.8)


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A freely falling object is an object that moves under the influence of gravity only.

Neglecting air resistance, all objects in free fall in the earth's gravitational field have a constant acceleration that is directed towards the earth's center, or perpendicular to the earth's surface, and of magnitude g = 9.8 m/s2.

If motion is straight up and down and we choose a coordinate system with the positive y-axis pointing up and perpendicular to the earth's surface, we describe the motion with Eq.(2.7), Eq.(2.8), Eq.(2.9) with a - g , x y.

Fundamentals of physics Freely Falling Bodies

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Equations of Motion for the 1-d vertical motion of an object in free fall:

v = v0 - gt y = v0t – ½gt2 v2 = v0

2 - 2gy

Note: Since the acceleration due to gravity is the same for any object, a heavy object does not fall faster than a light object.

Fundamentals of physics Freely Falling Bodies

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A car travelling at a constant speed of 30 m/s passes a police car at rest. The policeman starts to move at the moment the speeder passes his car and accelerates at a constant rate of 3.0 m/s2 until he pulls even with the speeding car.

Find a) the time required for the policeman to catch the speeder and b) the distance travelled during the chase.

Fundamentals of physics Problems 2.1

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We are given, for the speeder (s): vos = 30m/s = vs and as = 0 and for the policeman (p): vop = 0 and ap = 3.0 m/s2.

a)Distance travelled by the speeder xs = vst + ½apt2 = (30)t + ½(0)t = 30t. Distance travelled by policeman xp = vopt + ½apt2 = (0)t + ½(3.0)t2 = 1.5t2 When the policeman catches the speeder xs = xp or 30t = 1.5t2. 60t = 3t2 t2 – 20t = 0 t(t-20) = 0

Solving for t we have t = 0 or t = (20) = 20 s . The first solution tells us that the speeder and the policeman started at the same point at t = 0, and the second one tells us that it takes 20 s for the policeman to catch up to the speeder.

Fundamentals of physics Solution 2.1(a)

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We are given, for the speeder (s): vos = 30m/s = vs and as = 0 and for the policeman (p): vop = 0 and ap = 3.0 m/s2. We have found out the time taken = 20s.

(b) Substituting back in above we find, xs = vst + ½apt2 xs = 30(20) + ½(0)(20)2 = 600m

or, xp = vopt + ½apt2

xp = (0)(20) + (3.0)(20)2 = 600m.

Fundamentals of physics Solution 2.1(b)

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A car decelerates at 2.0 m/s2 and comes to a stop after travelling 25m. Find a) the speed of the car at the start of the deceleration and b) the time required to come to a stop.


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We are given: a = - 2.0 m/s2 vo = ? Vf = 0 m/s2

x = 25 m t = ?

(a) From vf2 = vo

2 + 2ax we have vo2 = vf

2 - 2ax vo

2 = 0 -2(-2.0)(25) = 100 m 2/s 2 or v0 = 10 m/s.

(b) From vf = v0 + at we have t = (vf - v0) = (- 10) = 5 s.

Fundamentals of physics Solution 2.2

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A stone is thrown vertically upward from the edge of a building 19.6 m high with initial velocity 14.7 m/s. The stone just misses the building on the way down. Find a)the time of flight and b)the velocity of the stone just before it hits the ground. c)The distance from the top of the building to the highest point reached. d)The total distance travelled by the stone.


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We are given: Vo = 14.7 ms-1 (velocity at A); a = -9.8 ms-2

x = -19.6 m (vector from the A to D)t = ?s; Vf = ? ms-1 (at D)

(a) From X = Vot + ½at2

-19.6m = (14.7ms-1)t + 1/2(-9.8ms-2)t2 -19.6 = 14.7t - 4.9t2 4.9t2 - 14.7t - 19.6 = 04.9(t2 – 3t – 4) = 0(t2 – 3t – 4) = 0(t – 4) (t + 1) = 0 t = 4s or -1sThe four seconds represents the time for the entire journey... A up to B (1.5 s); B to C (1.5s) then C to D (1s). The –1s represents the time it would have taken for the stone to go up to A if it was thrown up from the ground (clearly not the case).Note the vector quantities (the – sign represents the direction of the vector as up is +ve):X -19.6 m; a -9.8 ms2; vo 14.7 s-1.


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We are given: vo = 14.7 ms-1 (velocity at A); a = -9.8 ms-2 ; x = -19.6 m (vector from the A to D); t = ?s; Vf = ? ms-1 (at D)

Things that can go wrong...(1) You can reverse all the signs... That is not technically wrong: it means you are using a coordinate system which makes down +ve... (we usually use up as +ve)...(2) You can reverse X (displacement) only... This would lead to (t2 - 3t + 4 = 0) and so your equation has no real solutions (b2>4ac in a quadratic equation)... Understandable because the stone never gets to 19.6m above the building... B is only 11.025m above A.(3) You can reverse vo (velocity) only... And this would lead to (t2 + 3t - 4 = 0). It represents the stone being thrown downwards... solutions are -4s and 1 s... At first glance this seems to yield correct answers.


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Here the 1s represents the time it takes to get to the ground D from A when thrown downward and the -4s represents the time if the stone was thrown up from A, A up to B (1.5s), B down to C (1.5 s). At C it velocity would be |14.7| m s-1 in a downward direction (satisfying the initial condition of reversing vo) and C up to D (1s)

We are given: vo = 14.7 ms-1 (velocity at A); a = -9.8 ms-2

x = -19.6 m (vector from the A to D); and worked out t = 4s (at D); vf = ?ms-1

(b) From Vf = Vo + at Vf = 14.7ms-1 + (-9.8 ms-2) (4s) Vf = -24.5ms-1 Note: The negative sign indicates that the stone is moving downwards...

(c) From vf2 = vi

2 – 2as x = (vf

2 - vi2) / 2a

x = (02 – 14.72) / 2(-9.8) x = 11.025 m |xup| = 11.025 m

(d) Total distance travelled |xtotal| = |xup| + |xdown| xdown = (vf

2 - vi2) / 2a

xdown = 30.625 m |xtotal| = 11.025 m + 30.625 m |xtotal| = 41.65m [double check (2 X 11.025 + 19.6)]

Fundamentals of physics Solution 2.3 ------------------------------------------

mechanics unit 5: motion and forces 5.6 motion in one dimension - speed and velocity, acceleration

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