![MEC2047
Now find the distance traveled:
Ds0-30 = ò v dt = (1/5)(1/2) (30)2
= 90 m
Ds30-48 = ò v dt = [(-1/3) (1/2) (t – 48)2
]
= (-1/3) (1/2)(48 – 48)2
– (-1/3) (1/2)(30 – 48)2
= 54 m
s0-48 = 90 + 54 = 144 m
vavg(0-90) = total distance / time
= 144 / 48
= 3 m/s
48
30
Example 2](https://image.slidesharecdn.com/mec2047-03rectilinearmotion3-250303124827-955ba561/85/MEC2047-LECTURE-03-Rectilinear-Motion-3-ppsx-16-320.jpg)
MEC2047 LECTURE 03+Rectilinear+Motion (3).ppsx
- 2. MEC2047 In many experiments, a velocity versus position (v-s) profile is obtained. If we have a v-s graph for the tank truck, how can we determine its acceleration along the trip? Applications
- 3. MEC2047 The velocity of a car is recorded from an experiment. The car starts from rest and travels along a straight track. If we know the v-t plot, how can we determine the distance the car traveled during the time interval 0 < t < 30 s or 15 < t < 25 s? Applications
- 4. MEC2047 Erratic Motion The approach builds on the facts that slope and differentiation are linked and that integration can be thought of as finding the area under a curve. Graphing provides a good way to handle complex motions that would be difficult to describe with formulas. Graphs also provide a visual description of motion and reinforce the calculus concepts of differentiation and integration as used in dynamics.
- 5. MEC2047 s-t Graph Plots of position vs. time can be used to find velocity vs. time curves. Finding the slope of the line tangent to the motion curve at any point is the velocity at that point (or v = ds/dt). Therefore, the v-t graph can be constructed by finding the slope at various points along the s-t graph.
- 6. MEC2047 Also, the distance moved (displacement) of the particle is the area under the v-t graph during time t. Plots of velocity vs. time can be used to find acceleration vs. time curves. Finding the slope of the line tangent to the velocity curve at any point is the acceleration at that point (or ). Therefore, the acceleration vs. time (or a-t) graph can be constructed by finding the slope at various points along the v-t graph. v-t Graph
- 7. MEC2047 Given the acceleration vs. time or a-t curve, the change in velocity (v) during a time period is the area under the a-t curve. So we can construct a v-t graph from an a-t graph if we know the initial velocity of the particle. a-t Graph
- 8. MEC2047 A more complex case is presented by the acceleration versus position or a-s graph. The area under the a-s curve represents the change in velocity (recall ). This equation can be solved for v1, allowing you to solve for the velocity at a point. By doing this repeatedly, you can create a plot of velocity versus distance. a-s Graph = area under the a-s graph
- 9. MEC2047 Another complex case is presented by the v-s graph. By reading the velocity v at a point on the curve and multiplying it by the slope of the curve (dv/ds) at this same point, we can obtain the acceleration at that point. Recall the formula Thus, we can obtain an a-s plot from the v-s curve. v-s Graph
- 11. MEC2047 Example 1 A sports car is moving along a straight road with the s-t graph shown. Find the v-t graph and a-t graph over the time interval shown.
- 12. MEC2047 The v-t graph can be constructed by finding the slope of the s-t graph at key points. What are those? when 0 < t < 5 s; v0-5 = ds/dt = d(3t2 )/dt = 6 t m/s when 5 < t < 10 s; v5-10 = ds/dt = d(30t75)/dt = 30 m/s v-t graph v(m/s) t(s) 30 5 10 Example 1
- 13. MEC2047 Similarly, the a-t graph can be constructed by finding the slope at various points along the v-t graph. Using the results of the first part where the velocity was found: when 0 < t < 5 s; a0-5 = dv/dt = d(6t)/dt = 6 m/s2 when 5 < t < 10 s; a5-10 = dv/dt = d(30)/dt = 0 m/s2 a-t graph a(m/s2 ) t(s) 6 5 10 Example 1
- 14. MEC2047 For the 0 - 48 s interval on the v-t graph shown, find a) the a-t graph, b) the average speed, c) and distance traveled. Example 2
- 15. MEC2047 Plan: Find slopes of the v-t curve and draw the a-t graph. Find the area under the curve. It is the distance traveled. Finally, calculate average speed (using basic definitions!). Find the a–t graph: For 0 ≤ t ≤ 30 a = dv/dt = 0.2 m/s² For 30 ≤ t ≤ 48 a = dv/dt = -0.333 m/s² a-t graph -0.33 0.2 a(m/s²) 30 48 t(s) Example 2
- 16. MEC2047 Now find the distance traveled: Ds0-30 = ò v dt = (1/5)(1/2) (30)2 = 90 m Ds30-48 = ò v dt = [(-1/3) (1/2) (t – 48)2 ] = (-1/3) (1/2)(48 – 48)2 – (-1/3) (1/2)(30 – 48)2 = 54 m s0-48 = 90 + 54 = 144 m vavg(0-90) = total distance / time = 144 / 48 = 3 m/s 48 30 Example 2
- 17. MEC2047 A particle starts from rest at x = -2 m and moves along the x- axis with the velocity history shown. Plot the corresponding acceleration and the displacement histories for the 2 seconds. Find the time t when the particle crosses the origin. Example 3
- 19. MEC2047 The position coordinate of a particle which is confined to move along a straight line is given by , where s is measured in meters from a convenient origin and t is in seconds. Determine (a) the time required for the particle to reach a velocity of 72 m/s from its initial condition at t = 0, (b) the acceleration of the particle when v = 30 m/s, and (c) the net displacement of the particle during the interval from t = 1 s to t = 4 s. Example 4