Course Name : Physics I Course # PHY 107. Note - 3 : Motion in One Dimension

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1 Course Name : Physics I Course # PHY 107 Note - 3 : Motion in One Dimension Abu Mohammad Khan Department of Mathematics and Physics North South University Copyright: It is unlawful to distribute or use this lecture material without the written permission from the author

2 Topics to be studied Definitions of Physical Quantities Geometrical Interpretation of instantaneous and average velocities Geometrical Interpretation of Acceleration Equations of Motions Integral form of Displacement and Velocity Free Fall: Definition and Properties Examples Suggested Problems

3 Definitions of Physical Quantities: One Dimension means that there is only one coordinate, and conventionally it is the x-axis (horizontally) or y-axis for vertical motion (like free-fall motion, to be covered later on). The direction of a vector quantity is given by the sign of the parameter whether it is positive or negative. By convention, to the right is positive and to the left is negative directions. Similarly, upward is positive and downward is negative. Definitions of physical quantities: 1. Position: x(t) = Coordinate x as function of time t = (t,x(t)). 2. Displacement, x x(t) x(t 0 ) = x x 0, where t 0 is the initial time, and t is any { later time. Length of the Path 3. Distance, d = x (Magnitude of the Displacement).

4 4. Instantaneous velocity (or simply velocity): It is defined as, v(t) dx(t) = dx dt dt = First Derivative of the Position curve 5. Average velocity: It is defined as v av = x t = x x 0 (Definition) t t 0 = Slope of the straignt line that passes through the points (t,x) and (t 0,x 0 ) 6. Speed: It has two meanings defined by v av = Magnitude of the Average velocity v = d Length of the Path = t Time

5 7. The instantaneous acceleration is defined as a(t) = dv = First derivetive of the velocity graph at time t dt = Slope of the tangent line on the velocity graph at time t = d2 x dt 2 = Second derivative of the position graph at time t = Curvature of the position graph at time t Note that the acceleration is NOT constant in general, and is a function of time. It just gives how fast the velocity itself is changing. If the velocity does not change (means constant), the a = 0. DO NOT CONFUSE ACCELERATION WITH VELOCITY. THESE ARE COMPLETELY DIFFERENT!!! Note that, in general, the average velocity between the points b and e is not equal to the average velocities at points b and e, i.e. v av = x t = x e x b v b +v e t e t b 2.

6 The position vs. time graph: x(t) a b v b = Slope of the Tangent line at point b (dotted line) v av = Slope of the line = Average velocity between points b and e c d v = Slope of the Tangent line e at point e (dotted line) e t Note the following: v a = v c = v e = 0, because the slope of the dotted line is zero at these points. The curvature at point a is positive, because a a = d2 x dt 2 > 0. But the curvature at d is zero, because the line is straight through the point d.

7 Some properties of motion: Based on the mathematical properties of the first and second derivative, it can be concluded that: Speeding up implies that either (v > 0, a > 0) or (v < 0, a < 0). That is, both velocity and acceleration must have the same sign, either both positive or both negative. Slowing down implies that either (v > 0, a < 0) or (v < 0, a > 0). That is, velocity and acceleration must have opposite signs. Rest (actual rest) means that both v = 0 AND a = 0. Momentarily at rest (also known as the turning point ): This requires that ONLY v = 0, BUT a 0.

8 x(t) a b c d t Note the following: Note that the points a and c are turning point, because velocities are zero, but acceleration are not zero. The velocities changes directions at the turning points, from positive to negative or negative to positive directions. From a b, it is speeding up, because velocity and acceleration are positive. From b c, it is slowing down, because the velocity is positive, but the acceleration is negative. Again from c d, it is speeding up, because both the velocity and the acceleration are negative.

9 Equations of Motion: Constant acceleration: Since a=constant, the position x must be a quadratic function of time t, or linear function of time. The position x can not be a function of time with negative powers. The position can also not be a function of time with fractional powers. Otherwise the acceleration, being a double derivative of position with respect to time, will not be a constant. Note the following: Let s choose the initial time equals zero, i.e. t 0 = 0. From the definition of average velocity, we find: x = x 0 +v av t = x 0 +vt. From the definition of average acceleration, we also find, v = v 0 +a av t = v 0 +at = v 0 +at. Note that for constant acceleration, a av a = a. Using v = (v +v 0 )/2 which is the arithmetic mean of velocities, we also get, x = x 0 +v 0 t at2. Finally, removing time t from the previous expression, we also get, v 2 = v a(x x 0).

10 Short Summary: Only for constant acceleration, the equations of motions are: (1) x = x 0 +vt. (2) v = v 0 +at. (3) x = x 0 +v 0 t at2. (4) v 2 = v a(x x 0). Only for constant acceleration, the average velocity can also be expressed as the arithmetic average of two velocities at the final and initial times, i.e. v = v av = x t v +v 0 2 (Definition. So always valid) ((Valid iff acceleration is constant)

11 Integral Form of Displacement and Velocity: From the definition of velocity, we write: v = dx dt dx = vdt Integrating : t x x 0 dx = t t 0 vdt, x x x 0 = vdt = Area (A) under the curve in v-t graph t 0 = v t (iff v is constant) Note in the v-t graph that: A 1 > 0, A 2 > 0, A 4 > 0 and A 3 < 0. Therefore, the displacements are: v(t) Areas are positive above the t axis x b x a = A 1, x c x b = A 2 +A 4 A 3, a A 1 b A 2 c A 4 and x c x a = (A 1 +A 2 +A 4 ) A 3. t A 3 Area is negative below the t axis

12 The Acceleration vs. Time Graph: Similarly, from the a-t graph, it is easy to see that the areas are: A 1 > 0, A 2 > 0 and A 3 < 0. Therefore, the changes in velocity are: a(t) A 1 A 2 v b v a = A 1, v c v a = A 1 +A 2, and v d v a = (A 1 +A 2 ) A 3. a b c d A 3 t Note that when the acceleration is constant, the graph is horizontal. So, the region under the curve is perfectly rectangular, and the area under the curve is (a)( t), where a is the height of the rectangle and t is the width of the rectangle.

13 Free Fall: The free fall is an example of one-dimensional motion. The position is represented by the vertical axis (the y-axis). So, the notation has been changed accordingly. The conditions to be satisfied for any Free-Fall motion are: 1. The acceleration is given by: a = g = 9.80m/s 2. Therefore, g = a = 9.80m/s The acceleration is fixed (both in magnitude and direction) by the Gravitational force, NOT by any other means. Therefore, the equations of motion for Free-Fall become: (1) y = y 0 +vt. (2) v = v 0 gt. (3) y = y 0 +v 0 t 1 2 gt2. (4) v 2 = v 2 0 2g(y y 0).

14 Free-Fall Properties: The time to go up equals the time to go down if the displacement is zero: t up = t down = v 0 g (v 0 = initial speed) At any level ( i.e. same height from the ground): Max. Height (Turning point) v=0, a = g v up = v down = v up = v down. v up v down The total time of the flight (when displacement equals zero) is T = 2t up = 2t down = 2v 0 g. At the max. height, the velocity changes direction, but the acceleration is still non-zero constant. Hence, it is a turning point. t up t down v 0 v= v y 0 = y (Ground)

15 Example: The position of an object moving along an x axis is given by x = 3t 4t 2 +t 3, where x is in meters and t in seconds. Answer the following: (a) The object s displacement between t = 0 to t = 4s.; (b) What is its average velocity for the time interval from t = 2s to t = 4s? (c) Find the turning point of the object. (d) When will the acceleration be zero?

16 Solution: (a) The displacement is x = x 4 x 0 = (b) The average velocity is [ ( ) (0)] m = 12m. v av = x t = x 4 x 2 = ( ) ( ) m/s = 7m/s. t 4 t (c) At the turning point: v = 0. Therefore, v = dx/dt = 3t 2 8t +3 = 0. The solution is t = ( 8)± ( 8) sec = 0.45sec or2.2sec. 2 3 These are when the turning points occur. The location (or position) where the turning points occur are: x(t = 0.45) = [ 3(0.45) 4(0.45) 2 +(0.45) 3 ) ] m = 0.63m, and x(t = 2.2) = [ 3(2.2) 4(2.2) 2 +(2.2) 3 ) ] m = 2.1m. (d) The acceleration is: a = dx2 = 8+6t. Therefore, a = 0 yields t = 1.33sec dt2

17 Problem # 2.34 In the figure below, a red car and a green car, identical except for the color, move toward each other in adjacent lanes and parallel to an x axis. At time t = 0, the red car is at x r = 0 and the green car is at x g = 220m. If the red car has a constant velocity of 20km/h, the cars pass each other at x = 44.5m, and if it has a constant velocity of 40km/h, they pass each other at x = 76.6m. What are: (a) the initial velocity and (b) the constant acceleration of the green car?

18 Solution # 2.34 The schematic diagram is shown in the adjacent figure. Here, v 0 and a are unknown. In the Case-I, we need to find t for the red car, and write an equation for the green car (because t is same for both cars). In the Case-II, again we find t for the red car, and write an equation for the green car (same reason as before, but for different distance). Thus, we will get two equations with two unknown and solve by using the method of elimination. Case I: v=20km/h = 50 9 m/s t t t=0 x = 44.5 m (Cars meet here after time t) Case II: 100 v=40km/h = m/s 9 t=0 t x = 76.6 m (Cars meet here after time t) t v 0 a v 0 a

19 Case-I: Suppose after time t the cars meet at a position ( or distance) 44.5m. Since, v = 20km/h = 50/9m/sec, the time t is t = x v = sec = 8.0sec. 50 Now, in time t = 8.0sec, the green car s position satisfy (3rd equation of the motion) 44.5 = 220+8v a(8)2 = 32.0a+8.0v =0. Case-II: In this case, the cars meet at a position (or distance) 76.6m. Since, v = 40km/h = 100/9m/sec, the time t is t = x v = sec = 6.9sec. Now, 100 in time t = 6.9sec, the green car s position satisfy (3rd equation of the motion) 76.6 = v a(6.9)2 = 23.8a+6.9v = 0. Solving these equations (red colored equations) for initial velocity and acceleration using the method of elimination, we find that Initial velocity of the green car, v 0 = 13.5m/s. The acceleration of the green car, a = 2.1m/s 2.

20 Problem: A baseball is tossed up with initial speed of 12m/s. (a) How long does it take to reach the maximum height? (b) Find the maximum height. (c) How long dies it take to reach 5.0m above the release point? (d) What is the velocity at 5.0m above the release point? Solution: Here the ground is the release point: y 0 = 0. (a) Here v = v 0 +at gives: 0 = t up t up = (12/9.8)sec = 1.22sec. (b) In this case, we have, v 2 = v0 2 +2a(y y 0) = 0. Therefore, we find, h = y y 0 = v2 0 2g = (12)2 m = 7.35m t up Max height (v=0) 5.0 m v h v 0 = 12 m/s y 0 =0

21 (c) Here y = 5.0m. Therefore, the 3 rd equation of motion gives, y = y 0 +v 0 t at2, 5 = 0+12t 1 2 (9.8)t2, Max height (v=0) 0 = 4.9t 2 12t +5, t = ( 12)± ( 12) sec, = 0.53sec, 1.91sec. Note that 0.53sec is the time when the velocity is positive (that is it, going upward), and 1.92sec is the time when the velocity is negative (that is, going downward). t up h 5.0 m v v 0 = 12 m/s y 0 =0

22 (d) In the previous part, we found two times. That is the baseball was at the height of 5m from the ground twice: at 0.53sec and 1.92sec. The velocities at these two times are: v(t = 0.53sec) = ( v 0 +at ) t=0.53sec, = ( ) m/s, = 6.81m/s. ( ) v(t = 1.92sec) = v 0 +at, t=1.92sec = ( ) m/s, = 6.81m/s. t up Max height (v=0) 5.0 m v h v 0 = 12 m/s y 0 =0

23 Suggested Problems: Chapter # 3: 33, 34, 36, 43, 44, 45, 47 and 63.

Course Name : Physics I Course # PHY 107

Course Name : Physics I Course # PHY 107 Notes-3 : Motion in One Dimensions Abu Mohammad Khan Department of Mathematics and Physics North South University http://abukhan.weebly.com Copyright: It is unlawful