Linear Motion Review

Level : AP Physics Teacher : Kim

1-Dimension(=linear) A jogger starts at position A and runs in a straight line, stops at 10m, reverse direction to position B. A B

0m 6m 10m - The (total) distance the jogger traveled is 10m+4m=14meters - The displacement(=change in position) of the jogger is 6meters.

2-Dimension ex) A jogger starts at position A and runs to position B. The dotted line is the actual path of the jogger

A●

4m 5m

●B

- The (total) distance the jogger traveled is 5m + 4m + 5m = 14meters - The displacement(=change in position) of the jogger is 10.8meters ( √102 + 42=10.8meters

풕풐풕풂풍 풅풊풔풕풂풏풄풆 ■ Average Speed (v) for linear motion is defined as v = 풕풐풕풂풍 풕풊풎풆 푡표푡푎푙 푑푖푠푡푎푛푐푒 ex) If the jogger took 10 seconds to run from A to B, then the average speed is v = = 푡표푡푎푙 푡푖푚푒 14 =1.4m/s 10

휟퐱 풙−풙 ■ Average Velocity(v) for linear motion is defined as v = = 풐 휟풕 풕−풕풐 where v is velocity, Δx is displacement and Δt is time interval 휟퐱 10.8 Ex) If the jogger took 10 seconds to run from A to B, then the average velocity is v = = =1.08m/s 휟풕 10

휟풙 풅풙 ■ Instantaneous Velocity or Speed vx ≡ 풍풊풎 = ( linear) 휟풕→ퟎ 휟풕 풅풕 *~For linear motion, instantaneous velocity and instantaneous speed is similar. Instantaneous speed is just how fast an object is moving regardless of the direction, while instantaneous velocity represents how fast and direction. ■ Acceleration

휟풗 풗−풗풐 훥푣 풅풗 Average acceleration is defined at a = = and instantaneous acceleration is ax ≡ 푙푖푚 = 휟풕 풕−풕풐 훥푡→0 훥푡 풅풕 If an object changes its speed in a given time, there is acceleration. If the acceleration is constant, that is, if the object is increasing or decreasing its speed at a 'steady' rate, then we can use the formulas below. ex) The following snapshot is an object with constant acceleration. Notice that the object’s change in position increases at a steady rate

t=0 t=1s t=2s t=3s ex) The following snapshot is an object when the acceleration is not constant. Notice that speed of the object speeds up and slows down erratically.

t=0 t=1s t=2s t=3s t=4s t=5s t=6s

The following formulas can only be used when the acceleration ‘a’ is constant !!

2 2 2 v = vo+ at x – xo = vot + (½)at v = vo +2a(x – xo) vo : initial velocity, v : final velocity, xo : initial position, x : final position

■ Warm-up Questions *~assume acceleration is constant for all questions 1. A car travels a distance of 400 meters in 20 seconds at constant speed. What is the speed of the car? (The car is moving at a straight line) (a) 6m/s (b) 12m/s (c) 16m/s (d) 20m/s

2. If you run at a constant speed of 5m/s for 5 seconds, how many meters did you cover? (Assume you are running in a straight line) (a) 25m (b) 16m (c) 12m (d) 3m

3. A runner travels 1.5 laps around circular track in a time of 50s. The diameter of the track is 40m and its circumference is 126m. i) Find the average speed of the runner (a) 2.55m/s (b) 3.78m/s (c) 5.25m/s (d) 7.43m/s

ii) Find the runner’s average velocity (a) 3.52m/s (b) 1.25m/s (c) 0.80m/s (d) 0.21m/s 2 2 2 v = vo+ at x – xo = vot + (½)at v = vo +2a(x – xo)

4. A jogger starts from rest and speeds up with an acceleration of 3m/s2 in 2 seconds, what is the speed at 2 seconds? (a) 3m/s (b) 5m/s (c) 6m/s (d) 14m/s

5. Car-1 is moving at a constant speed of 30m/s. At the instant car-1 pass car-2, car-2 starts from rest and speeds up to 30m/s in 10 seconds. Which car traveled further in 10 seconds?

6. A body with initial velocity 8m/s moves along a straight line with constant acceleration and travels 640m in 40s. For the 40s interval, i) find the average velocity a) 16m/s b) 12m/s c) 6m/s d) 0.6m/s

ii) find the final velocity, that is the velocity when the time is t=40s. (you need to find the acceleration first and then solve for final velocity) a) 29m/s b) 24m/s c) 19m/s d) 12m/s

7. A rocket-propelled car starts from rest and moves in a straight line with a=5m/s² for 8 seconds until the fuel is exhausted. It then continues with constant velocity. What distance does the rocket-car cover in 12 seconds? a) 160m b) 260m c) 320m d) 540m

2 2 2 v = vo+ at x – xo = vot + (½)at v = vo +2a(x – xo)

8. A box slides down an incline with uniform acceleration. It starts from rest and attains a speed of 2.7m/s in 3s. i) Find the acceleration a) 7.25m/s2 b) 4.51m/s2 c) 2.13m/s2 d) 0.90m/s2

ii) Find the distance moved in the first 6s a) 1.20m b) 5.9m c) 16.2m d) 21.8m

Signs of the velocity and acceleration 'a' If a certain direction is chosen as positive, the sign of the velocity in that direction will be positive whether the object is speeding up or slowing down. If the object reverse direction, then the velocity will be negative whether the object is speeding up or slowing down.

If ‘a’ is positive(a>0), then the acceleration is in the positive x-direction(right). If ‘a’ is negative(a<0), then the acceleration is in the negative x-direction(left).

Determine the signs of ‘a’ for the following cases. We choose right as positive direction.

(a) If an object is moving to the right and the speed is increasing (b) If an object is moving to the right and the speed is decreasing (c) If an object is moving to the left and the speed is increasing (d) If an object is moving to the left and the speed is decreasing

9. An object is moving with uniform(=constant) acceleration has a velocity of 12m/s in the positive x- direction when its x-coordinate is 3m. i) If its x-coordinate 2s later is –5m, what is its acceleration? a) 16m/s2 b) –16m/s2 c) 12m/s2 d) –12m/s2

ii) Find the velocity at 2s a) 14m/s b) –14m/s c) 20m/s d) –20m/s

iii) Find the time when the object momentarily stops. a) 0.25s b) 0.50s c) 0.75s d) 1.00s