advertisement

Topic 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 Displacement Vectors Kinematics Graphs Energy Power Springs Shadows Field of Vision Colors Concave mirrors Convex mirrors Refraction Lenses Optical Power Slides Minutes 9 27 13 39 13 39 10 30 10 30 5 15 4 12 3 9 7 21 3 9 7 21 4 12 5 15 10 30 6 18 Distance-Time Graphs Slope = m / s = velocity Area = m x s = ms = nothing 8 s (m) Distance-Time Graph 7 6 Slope = velocity Area = nothing 5 4 3 2 1 t(s) 0 1 a) b) c) d) Example At At At At 2 3 4 5 6 7 8 9 10 11 11 12 13 14 t = 2 s velocity = 1 m/s (slope = 1) t = 5 s velocity = 0 (slope = 0) t = 9 s velocity = -1 m/s (slope = -1) t = 12 s velocity = 0 (slope = 0) Click to continue Distance –vs- Time graph In a distance versus time graph, the slope represents the velocity. And the area under the curve, does not represent anything. 80 s (m) 70 60 50 40 30 20 Area = nothing 10 t (s) 0 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 To find the velocity at any point, find the slope at that point. Click Velocity-Time Graphs In a velocity versus time graph, the slope represents the acceleration. And the area under the curve, represents the distance traveled. 80 v (m/s) 70 60 50 40 30 20 Area = distance traveled 10 t (s) 0 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 To find the acceleration at any point, find the slope at that point. To find the distance traveled between any two points, find the area under the curve between those two points. Click 8 v (m/s) 7 Velocity-Time Graph 6 Slope = acceleration Area = distance 5 The graph on the left illustrates the velocity-time curve of a vehicle. 4 Find the total distance traveled by the vehicle. 3 2 C 1 A B D t(s) 0 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 Area of sector A = (3 m/s x 3 s)/2 = 4.5 m Area of sector B = 3 m/s x 5 s = 15 m There arewe a number ways oftime, finding the total area under curve. Since First velocity find is plotted theCof area versus of each segment we know that then theadd slope up aall represents the Area of sector = (3 m/s x 2 s)/2 =2 mand Inacceleration the example above, one way is to divide the total into To the total distance, we need to the find the total area under thefour curve. segments obtain the total area and thus thearea answer. thefind oftothe vehicle and area under the curve represents Area of (segment-A, sector D = 1 segment-B, m/s x 6 s = 6segment-C m sections and segment-D). the distance traveled by the vehicle. Distance traveled = A + B + C + D = 4.5 m + 15 m + 2 m + 6 m = 27.5 m Example-1 Click to continue 8 7 v (m/s) Velocity-Time Graph 6 The graph on the left illustrates the velocity-time curve of a vehicle. Slope = acceleration Area = distance 5 4 Find the distance traveled by the vehicle for the first 5 s. 3 2 1 A B t(s) 0 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 Answer Step-1: Area of sector A = (3 m/s x 3 s)/2 = 4.5 m Note that the first 5for seconds is5 from t =curve 0 tooft two = 5 s. Thus,Since we need the to distance find the distance the firstunder s consists the from segments, t = 0 to t = 5 s. Step-2: Area of sector A = (3 m/s x 2 s) = 6 m we will find the area in two steps. Step-1 is from t = 0 to t = 3 s (segment A) and step-2 is from t = 3 s to t = 5 s (segment B). Distance traveled = A + B = 4.5 m + 8 m = 10.5 m Example-2 Click to continue 8 7 v (m/s) Velocity-Time Graph 6 The graph on the left illustrates the velocity-time curve of a vehicle. Slope = acceleration Area = distance 5 Find the distance traveled by the vehicle between t = 4 s and t = 7 s. 4 3 2 1 t(s) 0 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 Note that Area the under area under the curve the curve = 3 m/s for xthis 3 sproblem = 9 m is from t = 4 s to t = 7 s Thus, all we need to do is find the area of the rectangle. and consists of only one section. Distance traveled = 9 m Example-3 8 7 v (m/s) Velocity-Time Graph 6 The graph on the left illustrates the velocity-time curve of a vehicle. Slope = acceleration Area = distance 5 Find the distance traveled by the vehicle during the last 5 seconds. 4 3 2 1 t(s) 0 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 1 m/s by 1 s 1 m/s by 5 s The area under the curve for this problem consists of two sections, Note that the area under the curve for this problem is from t = 9 s to t = 14 s. a small Answer Area of triangle = (1triangle m/s x 1and s)/2a =rectangle. 0.5 m Area of rectangle = (1 m/s x 5 s) = 5 m Distance traveled = 0.5 m + 5 m = 5.5 m Example-4 Graphs Slide: 4. 1 During the course of a laboratory experiment, a team of students obtained the graph on the right representing the force exerted as a function of the acceleration of a cart. Calculate the mass of the cart. A) 3.0 kg B) 2.0 kg C) 1.0 kg D) 0.50 kg E) 0.25 kg Note that the slope represents mass. Graphs Slide: Reminder The area under the curve is the distance travelled. 4. 2 200 m 250 m 200 m + 100 m Answer: 350 m – 200 m = 150 m 350 m Click Graphs Slide: 4. 3 Illustrated below is the velocity versus time graph of a particle. NOTE: The area under the curve represents distance. 250 000 50 000 200 000 How far has the particle travelled in 5 minutes? A) 200 000 m B) 250 000 m C) 300 000 m D) 350 000 m Convert to seconds 5 min = 5 x 60 s = 300 s Graphs Slide: 4. 4 Consider the position versus time graph below. Which one of the following statements best describes the motion illustrated by the above graph. A) Increasing velocity, constant velocity, increasing velocity B) Increasing velocity, zero velocity, increasing velocity C) Constant velocity, constant velocity, constant velocity D) Constant velocity, zero velocity, constant velocity Click Zero acceleration Positive acceleration Graphs Slide: 4. 5 The velocity-time graph on the left represents the motion of a car during a 6 s interval of time. Negative acceleration A) B) Which of these graphs represents the acceleration of the car? C) D) Click Graphs Slide: 4. 6 Fastest velocity here thus maximum KE at this point. This point shows that the car has zero KE which means it has zero velocity at point I which is incorrect. Click Graphs Slide: 4. 7 Velocity = 750 m/25 s = 30 m/s Area = velocity = 50 m/s Velocity = 600 m/25 s = 24 m/s Velocity = 25 m/s Click Slide: 4. 8 This means second segment has a negative This the means the first segment has to be aacceleration straight line (slope). (a=0). Click Graphs Slide: 4. 9 Since the distance between dots is increasing, the object is accelerating. Zero velocity Constant forward velocity Constant reverse velocity Click Graphs Slide: 4. 10 Carlo Martini owns a Ferrari. Carlo performed a speed test on his car and plotted the graph below. Note that the slope represents acceleration. Knowing that Carlo’s Ferrari has a mass of 1288 kg, calculate the net force of the car based on the above graph. Click Graphs Slide: 4. 13 Both a car and a truck drive off at the same time and in the same direction. The graphs below illustrate their movement. Area to x-axis = distance Slope = Average velocity = 10 m/s Determine how far apart the two vehicles are after 30 seconds. Step-1 Distance of car = area under the curve = 275 m Step-2 Distance of car = VAt = (10 m/s)(30 s) = 300 m Step-3 Distance apart = 300 m – 275 m = 25 m Click … and good luck!