The area under the velocity/time curve is equal to the total change in displacement

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Mousetrap.car.notes.problems Topics that will be studied with mousetrap cars are: motion in one dimension under constant acceleration torque and its relationship to angular and linear acceleration angular quantities of motion moment of inertia and its relationship to torque and angular acceleration energy losses due to friction energy stored in a spring Displacement. This is a vector and is defined as the final point an object is at minus the initial point. Another way to say it is the change in position of an object. Displacement must indicate or at least imply a direction. Distance. This is a scalar Average Velocity This is a vector and is defined as the change of displacement divided by the time taken. Average speed This is a scalar quantity and is equal to distance divided by time. Instantaneous velocity. This is the velocity at a given moment. On a graph of displacement versus time, the instantaneous velocity is the slope of the curve at a given point on the graph. The area under the velocity/time curve is equal to the total change in displacement acceleration This is a vector (there is no corresponding scalar quantity to acceleration). It is defined as the change of velocity divided by the change in time or time taken For the graph of velocity vs. time the instantaneous acceleration is the slope of the curve at any given point Newtons equations of motion* The big daddy of them all

d = d o + v o t + 1 2 at 2 the little o is pronounce sub-naught and stands for initial. d is displacement or distance and sometimes is replaced by s or h (in this case it means height when the formula is used for vertical motion. In this case a is replaced by g and made negative since it points down d = 1 2 at2 A simplification of the above where do= 0 and the object starts from rest. d = vt if there is no acceleration and the object starts from d= 0, this is a simplification of the above equation v 2 = v o 2 + 2ad This is one of the handiest formulas because you do not need to solve for time. Make sure you know how to use it. We will use this to determine the negative acceleration on the car due to the forces of friction v average = 1 ( 2 v + v o final) Notice that this is just how you take the average of anything. We will use this formula in determining the velocity of our mousetrap cars just as the mousetrap spring plays out and releases. v = v o + at often when using this equation you can assume that v o is zero (the object starts from rest) 1 2 kx 2 = Energy stored in a spring k is called the spring constant or the force constant and is a measure of the stiffness of the spring. K is measured in newtons/meter F=-kx This known as Hooke s Law and show the linear relationship between how far a spring is stretched and the force that the spring exerts in the opposite direction (hense the negative sign). Force is in newtons, x must be in meters (it is usually given in cm and you must convert it) examples:

1. A mousetrap car accelerates from rest for 4.5 seconds. In this time the car travels 12.5 ft. (a) What is the distance in meters? (b) What is the acceleration of the car? (c) How do you know that the car is no longer accelerating (d) If the uncertainty of the the stopwatch is 0.1 s, what is the percent uncertainty of the time? (e) If the uncertainty of the measurement of the distance is 4.0 inches, what is the percent uncertainty of the distance measurement? (f) What is the total % uncertainty of the calculation of the acceleration? (g) What is the car s maximum speed? (This occurs just as the spring finishes accelerating the car) (h) What is the average speed during the acceleration? Answers " 1m % 12.5 ft$ ' = 3.81m (a) # 3.28 ft& (b) Use d = 1 2 at2 so a = 2d t 2 = 2 " 3.81m (4.5s) 2 = 0.38 m s 2 (c) since force = ma, once the spring plays out and is no longer pulling on the wheels the car is no longer accelerating. The mousetrap will usually snap when this happens 0.1s (d) 4.5s x100 = 2.2% 0.33 ft (e) 12.5 ft x100 = 2.64% (4 inches = 0.33 ft) (f) add the uncertainties in quadrature (like the two legs of a right triangle) but since you square the time you double the % uncertainty of the measurement of time

total% _ uncertaint y = (2 " 2.2) 2 + (2.7) 2 = 5.2% Uncertainty of time uncertainty of distance. since the distance is squared to find acceleration you need to double this % uncertainty The uncertainties are added in quadrature because they are independent of each other (one uncertainty does not affect the other) (g) You can use either v = v o + at or v 2 f = v 2 o + 2ad where vo is 0 for both equations. vmax = 1.7 m/s (h) This is just 1/2 the max speed or 0.85 m/s 2. A spring has a constant k of 50 N/m. If it is stretched by a distance of 20 cm (a) How much energy is stored in the spring 1 2 kx 2 = 1 2 50 N m (0.20m)2 =1.0J (b) What is the restoring force exerted on the person pulling the spring and in what direction does it point? F=-kx F=50N/m 0.20m = 10 N in the opposite direction from which it was stretched 3. A spring of natural length 15 cm is hung from a nail in the ceiling. A mass of 5.0 kg is hung from the end of the spring and the new length of the spring is 25 cm. What is the k, the spring constant of the spring. x (elongation of the spring = 10 cm = 0.10 m) Force = mg = 5kg 9.8m/s 2 = 49 N Thus k = F/x 49N/0.10m = 490 N/m Homework 1. A mousetrap car accelerates from rest for 5.5 seconds. In this time the car travels 19.2 ft. (a) What is the distance in meters?

(b) What is the acceleration of the car? (c) How do you know that the car is no longer accelerating? (d) If the uncertainty of the the stopwatch is 0.1 s, what is the percent uncertainty of the time? (e) If the uncertainty of the measurement of the distance is 3.0 inches, what is the percent uncertainty of the distance measurement? (f) What is the total % uncertainty of the calculation of the acceleration? (g) What is the maximum speed of the car? (h) What is the average speed of the car during he acceleration? 2. A spring has a constant k of 75 N/m. If it is stretched by a distance of 30 cm (a) How much energy is stored in the spring (b) What is the restoring force exerted on the person pulling the spring and in what direction does it point? 3. A spring of natural length 15 cm is hung from a nail in the ceiling. A mass of 7.0 kg is hung from the end of the spring and the new length of the spring is 30 cm. What is the k, the spring constant of the spring. Answers: (a) 5.9 m (b) 0.39 m/s2 (c) The string is no longer pulling on the axle of the car (d) 1.8% (e) 1.3% (f) 3.18% (g) 2.1 m/s (h) 1.1 m/s 2. (a) 3.4 J (b) 23 N in the opposite direction 3. 229 N/m

Answers: (a) 5.9 m (b) 0.39 m/s2 (c) Ask your lab partner (d) 1.8% (e) 1.3% (f) 3.18% (g) 2.1 m/s (h) 1.1 m/s