Motion Part 2: Uniform Linear Motion

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1 Motion Part 2: Uniform Linear Motion Last modified: 13/03/2018

2 Links Motion Quantities Uniform Velocity in One Dimension Uniform Acceleration in One Dimension Notation Extra Equations Uniform Acceleration Problems Summary How to Solve Problems Example 1a Example 1b Example 1c Example 1d Acceleration Due to Gravity Example 2a Example 2b Example 3a Example 3b Example 3c Example 3d Example 4a Example 5a Example 5b Appendix: Derivations Using Calculus

3 Motion Quantities: Recap In the last lecture we introduced the THREE different quantities required to completely describe the motion of a point particle, each of them a function of time, t. POSITION, r(t) the location of the particle in the chosen reference frame. VELOCITY, v(t) = r t how quickly the position is changing ACCELERATION, a(t) = v t how quickly the velocity is changing At least in theory, if we know any one of these three functions then we can calculate the other two. The vectors r, v and a will, of course, in general be three-dimensional vectors.

4 Uniform Velocity in One Dimension The simplest possible motion is when velocity is uniform (i.e. unchanging, constant). In this case, the acceleration will be zero, and the motion will be in one dimension only. r r(t 2 ) r(t 1 ) r t V = r t The speed V is given by the gradient of the position-time graph (at left). The displacement s in a time interval t is given by the area under the velocity-time graph (at right): v V t t 1 t 2 t s = V t t 1 t 2 t

5 Uniform Acceleration in One Dimension When the acceleration is non-zero, but constant in the same direction as velocity, then the motion will still be in only one dimension. v The acceleration a is given by the gradient of the velocity-time graph (at left). v(t 2 ) v = v(t 2 ) v(t 1 ) = a t v(t 1 ) v t a = v t v(t 2 ) = v(t 1 ) + a t It is usually convenient to set the initial time t 1 = 0 and the final time t 2 = t, after which the above equation becomes: t 1 t 2 t v(t) = v(0) + a t

6 v v(t 2 ) v = a t v(t 1 ) t As seen previously, the change in displacement between two times can be determined by calculating the area under the velocitytime graph: s = v(t 1 ) t + 1 (a t) t 2 Again setting the initial time t 1 = 0 and the final time t 2 = t, this equation is: t 1 t 2 t s = v(0) t a t2

7 Notation Depending on which textbooks you look at, you may see several different notations used to write these equations more compactly: v(t) v, v(0) v 0. Using the subscript 0 to indicate t = 0 is a very common abbreviation which we will use later in the course. v(t) v f, v(0) v i. ( f for final and i for initial) v(t) v, v(0) u. These notes will use this last form. If you prefer to use one of the other notations that is absolutely fine.

8 Uniform Acceleration: Extra Equations Though these three equations are all we really need, two other equations can be derived simply from them: To express v in terms of s (i.e. eliminate t): v 2 = (u + a t) 2 = u 2 + 2uat + a 2 t 2 = u 2 + 2a (ut at2 ) = u 2 + 2as and s in terms of the velocities and time (eliminate a): s = ut at2 = ut (v u )t 2 t = 1 2 (u + v)t = v average t

9 Uniform Motion Equations The equations we need to solve uniform motion problems (a = constant) in one dimension are: v = u + at s = ut at2 v 2 = u 2 + 2as s = 1 (u + v)t 2 Many textbooks only give the first three of these, as the last equation is not actually that useful in practice.

10 Note that we have FIVE variables (a, u, v, s, t) and each of these equations connects FOUR of these variables. When deciding which equation to use it is sometimes helpful to focus on what we don t know. CAUTION: Unfortunately, people are a little inconsistent with notation with these equations. Normally, a refers to a and so is always positive. But here a (and also v, u, s) can be any value, positive or negative. The positive or negative value depends on the directions of the vectors. Though the equations don t contain any i,j or k, direction is still important.

11 How to Solve Problems These problems are all solved using the same basic steps, and should be straightforward as long as you are organized and careful. Read the problem carefully, and make a list of the information you are given. Draw a diagram, and pay particular attention to getting the correct directions. Read the question again, and check your diagram is correct. What are you asked to calculate? Find the formula that connects the known and required quantities. Do the calculation.

12 Example 1a A car accelerates uniformly from rest and after 5 s is travelling at 20 km/hr. What is the acceleration of the car? + rest u = 0 initial (t = 0) v = 20 km/hr= = 5.56 m/s a =? final (t = 5 s) Which formula links the given u,v,t with the required a? v = u + at a = (v u)/t = (5.56 0)/5 = 1.11 m/s 2 REMINDER: we must convert to SI units before calculating.

13 Example 1b How far did the car travel in the previous problem? + u = 0 s =? v = 5.56 m/s a = 1.11 m/s initial (t = 0) final (t = 5 s) Which formula links the known u,v,t,a with the required s? EITHER of s = ut at2 OR v 2 = u 2 + 2as will give the answer. s = ut at2 = = 13.9 m s = v 2 u 2 2a = = 13.9 m

14 Example 1c The same car, at the 5 s mark, brakes and accelerates uniformly to a halt after travelling a further 50 m. Calculate this acceleration. + s = 50 m u = 5.56 m/s a =? v = 0 initial (t = 0) final The previous final speed now becomes the initial speed. formula links the known u,v,s with the required a? Which v 2 = u 2 + 2as a = (v 2 u 2 ) 2s = (02 (5.56) 2 ) 2 50 = 0.31 m/s 2 NOTE: The negative sign indicates that the acceleration is opposite to the direction of motion - the car is decelerating.

15 Example 1d How long did the deceleration in the previous example take? + u = 5.56 m/s s = 50 m a = 0.31 m/s v = 0 t =? initial (t = 0) final As before, we have a choice of formulas which will work. The simpler one is: v = u + at t = (v u) a = (0 5.56) 0.31 NOTE: The negative sign of a is important. = 17.9 s

16 Acceleration Due to Gravity In the real world, cars generally don t accelerate uniformly, though this may sometimes be a useful approximation. A more realistic example is the case of objects falling due to gravity. The acceleration due to gravity is downwards and equal to g = 9.8 m/s 2. This is also an approximation (a very good one), as the value of g does vary slightly around the world, and of course decreases as we move into space. Often, rounding off to g = 10 m/s 2 is good enough. Unless an exam question specifies a value to use, either is acceptable. NOTE: The value of g is ALWAYS positive, even though the acceleration may be negative.

17 Example 2a A ball is dropped from a height of 20 m above the ground. What is its speed when it hits the ground? NOTE: The word dropped implies no initial motion. + rest, u = 0 a = g = 10 m/s 2 s = 20 m v =? The appropriate formula is: v 2 = u 2 + 2as = = 400 v = ±20 m/s Mathematically, there are 2 solutions, but clearly in this case v = +20 m/s is the correct choice.

18 Example 2b How long is the ball in the previous example in the air before it hits the ground? Again we have a choice of formulas to use: OR v = u + at t = v u a = = 2 s s = ut at2 20 = t2 t = ±2 s In the second case we again need to think carefully about which possible answer is the correct one.

19 Example 3a A ball is thrown vertically upward at a speed of 30 m/s. What is the maximum height reached by the ball? The first thing we need to realize is that the maximum height occurs when the ball momentarily stops, before falling back down, so v = 0. The required formula is: rest,v = 0 + v 2 = u 2 + 2as a = g = 10 m/s 2 s max =? u = 30 m/s 0 = ( 10)s = s s = = 45 m Note that, unlike the previous example, a is now negative.

20 Example 3b How long: (a) does it take the ball to reach the maximum height? (b) does it then take the ball to return to the ground? (c) is the ball in the air? (a) As we saw previously, at maximum height v = 0 so: v = u + at t = (v u)/a = (0 30)/( 10) = 3 s (b) We can calculate the time taken to fall to the ground from the maximum height of 45 m using the same method used in Example 2b: s = ut at2 45 = t2 t 2 = 9 t = ±3 s Note that the ball takes the same time to travel from the ground to the peak and then to return to the ground.

21 (c) Clearly the total time the ball is in the air ( time of flight ) is the sum of these two answers: = 6 s It is also possible to calculate this in a single step by realizing that the ball starts and finishes at ground level, so the displacement is zero (because the ball changes direction, distance travelled and displacement are different!). To find the time at which s = 0: s = ut at2 0 = 30t ( 10)t2 0 = 6t t 2 0 = t(6 t) t = 0 s OR t = 6 s In this case, t = 6 s is the required solution.

22 Example 3c At what speed does the ball hit the ground? We know u = 30 m/s and a = 10 m/s 2, and with the justcalculated value of t = 6 s, we find: v = u + at = 30 + ( 10)6 = 30 m/s The negative sign of course indicates that the motion is downward. If we hadn t already calculated the time, then we could calculate directly using s = 0 m: v 2 = u 2 + 2as = ( 10)(0) = 900 v = ±30 m/s And we know that the negative sign must be the correct one. Notice that the final speed is the same as the initial speed. The final and initial velocities are in opposite directions.

23 Example 3d For the ball in the previous question, at what time would it be at a height of 25 m? + t =? a = g = 10 m/s 2 s = 25 m u = 30 m/s The appropriate formula is: s = ut at2 25 = 30t ( 10)t2 0 = t 5t 2 0 = 5 6t + t 2 = (t 1)(t 5) t = 1 s or t = 5 s In this problem, both solutions are valid: one is on the way up, the other on the way down.

24 Example 4a A ball is thrown upward at 30 m/s from the edge of a 35 m high cliff. How long does it take for the ball to reach the ground at the base of the cliff? + u = 30 m/s The question can be rephrased as At what time t is s = 35 m? The appropriate formula is: a = g = 10 m/s 2 s = 35 m t =? s = ut at2 35 = 30t ( 10)t2 0 = t 5t 2 0 = 7 6t + t 2 = (t + 1)(t 7) t = 1 s OR t = 7 s

25 Clearly the correct answer in this case is t = 7 s. From previous calculations, we know that the ball reaches a height of 45 m before stopping and falling back down. So the distance travelled by the ball is: = 125 m This is a case where it is important to remember that distance displacement. This problem can also be done in steps: by calculating the time to reach the max height (3 s as calculated earlier), then the time to return to the top of the cliff (also 3 s) and finally the time to drop to the base of the cliff (1 s) and adding them together. Doing it with just one direct calculation is much neater though.

26 Example 5a A rocket s engine generates a constant acceleration of 30 m/s 2. The rocket is launched vertically, and after 10 s the engines run out of fuel. What is the maximum height reached by the rocket? We need to realize that in this problem, there are two different accelerations acting during the rocket s motion. We can t do the problem in one step - we must split it into two separate problems. + s =? t = 10 s, v =? a = 30 g = 20 m/s 2 The first stage of motion is when the engine is operating. To find s: s = ut at2 u = 0 = (20)102 = 1000 m

27 And to find the speed v when the engine stops: v = u + at = = 200 m/s This final speed is now the initial speed for the second phase of the motion, where we will calculate the height where the rocket (momentarily) comes to rest. + rest,v = 0 a = g = 10 m/s 2 v 2 = u 2 + 2as s =? 0 = ( 10)s u = 200 m/s s = = 2000 m The maximum height is then the sum of the values found for the two stages of the motion. i.e = 3000 m.

28 Example 5b For the rocket in the previous example, what is the total time of flight? In other words: how much time passes between the rocket taking off and returning to the ground? We already know that the first part of the motion takes 10 s. We now need to calculate the time taken for the second part and then add the two results. For the second part of the motion, the displacement is s = 1000 m u = 200 m/s + s = ut at2 a = 10 m/s 2 s = 1000 m t =? 1000 = 200t ( 10)t2 0 = t 2 40t 200 t = 44.5 s OR t = 4.5 s The negative answer can t be correct, so the total time must be: = 54.5 s

29 Calculus: 1st Uniform Motion Equation For uniform acceleration, velocity can be determined by a simple integration: v(t) = adt = at + c where c is a (vector) constant of integration, which we can determine by setting t = 0: v(0) = a 0 + c and so the expression for velocity at time t is: v(t) = a t + v(0)

30 Calculus: 2nd Uniform Motion Equation To calculate r(t), the position at time t: r(t) = v dt = (u + at)dt = u t a t2 + c where again, c is a (vector) constant of integration, that is determined by setting t = 0: r(0) = u 0 + a 0 + c and so our expression for position at time t is: r(t) = u t a t2 + r(0) and the displacement s is given by: s(t) = r(t) r(0) = u t a t2

Physics 101 Discussion Week 3 Explanation (2011)

Physics 101 Discussion Week 3 Explanation (2011) D3-1. Velocity and Acceleration A. 1: average velocity. Q1. What is the definition of the average velocity v? Let r(t) be the total displacement vector