The graph represents the motion of two cars. Which statement is correct? The blue car is always slower than the red car. The cars have the same speed at t = 12 sec. The cars have the same position at t = 12 sec. The blue car has greater acceleration. v(m/s) t (s) 0 1 2 3 4 5 6 7 8 9 10 2 4 10 16 14 12 6 8 18 20 Red Car Blue Car 11 12 13 14 15 16

Blue Car Red Car (b) The cars have the same speed at t = 12 sec. v(m/s) t (s) 0 1 2 3 4 5 6 7 8 9 10 2 4 10 16 14 12 6 8 18 20 Red Car Blue Car 11 12 13 14 15 16

A rock is dropped from a hovering helicopter A rock is dropped from a hovering helicopter. If it’s initial velocity is 0 m/s, and the acceleration due to gravity is 10 m/s2, what is the velocity of the rock after it has fallen for 5 seconds? 0 m/s 5 m/s 50 m/s 20 m/s v = ?

(c) 50 m/s t = 0 v = 0 t = 1 s v = 10 m/s t = 2 s v = 20 m/s Every second the rock falls, its velocity increases by 10 m/s. t = 3 s v = 30 m/s t = 4 s v = 40 m/s t = 5 s v = 50 m/s

From 0 – 6 sec, the graph shows speeding up in the positive direction. slowing down in the negative direction. speeding up in the negative direction. slowing down in the positive direction. t (s) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 -6 -4 2 8 6 4 -2 10 12 v(m/s)

(d) slowing down in the positive direction. The object’s motion begins with a velocity of 10 m/s and decreases at a constant rate to 0 m/s at t = 6 sec. t (s) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 -6 -4 2 8 6 4 -2 10 12 v(m/s)

From 0 – 6 sec, The velocity and acceleration are positive. The velocity and acceleration are negative. The velocity is positive, and acceleration negative. The velocity is negative, and acceleration positive. t (s) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 -6 -4 2 8 6 4 -2 10 12 v(m/s)

(c) The velocity is positive, and acceleration negative. The slope (acceleration) is negative, and the velocity is positive. t (s) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 -6 -4 2 8 6 4 -2 10 12 v(m/s)

From 6 – 11 sec, the graph shows speeding up in the positive direction. slowing down in the negative direction. speeding up in the negative direction. slowing down in the positive direction. t (s) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 -6 -4 2 8 6 4 -2 10 12 v(m/s)

(c) speeding up in the negative direction. The object’s motion begins with a velocity of 10 m/s and decreases at a constant rate to 0 m/s at t = 6 sec. t (s) 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 -6 -4 2 8 6 4 -2 10 12 v(m/s)

Which V vs. T graph corresponds to the X vs. T graph? x(m) t (s) v(m/s) t (s) (b) v(m/s) t (s) v(m/s) t (s) (a) (c)

(c) The slope is the velocity v(m/s) t (s) x(m) t (s)

Which V vs. T graph corresponds to the X vs. T graph? x(m) t (s) parabolic v(m/s) (b) t (s) v(m/s) v(m/s) t (s) (c) (a) t (s)

(b) Slope is velocity x(m) t (s) parabolic v(m/s) t (s)

A dog runs around a circular path as shown, starting at point P and stopping at point Q. What is the dog’s displacement? R = 50 meters R P Q N 2R, south R, southwest 100 meters, south R, north

(c) 100 meters, south The change in position is 100 m south P R Q R = 50 meters R P Q The change in position is 100 m south

What is the displacement from t = 0 - 8.0 sec?? 10 8 v (m/s) 6 4 2 t (s) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 -2 -4 -6 -8 0 meters 64 meters 1 meter 32 meters

(d) 32 meters, displacement is area. The Area beneath the graph is ½(8 m/s)(8 s) = 32 m v (m/s) t (s) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 -8 -6 6 4 2 -4 -2 8 10

What is the displacement from t = 0 - 16 sec?? 10 8 v (m/s) 6 4 2 t (s) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 -2 -4 -6 -8 0 meters 64 meters 1 meter 32 meters

(a) 0 meters v (m/s) t (s) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 -8 -6 6 4 2 -4 -2 8 10 A = 32 m A = -32 m

Identify time intervals of: Constant negative velocity Constant positive velocity Zero velocity

Identify time intervals of: Speeding up Slowing down Positive velocities Negative velocities

Identify time intervals of: Speeding up Slowing down Positive velocities Negative velocities

Find: Average speed for entire 20 sec Average velocity for t = 8 – 20 sec Average velocity for entire 20 sec Instantaneous velocity at t = 16 sec

Find: Average acceleration for t = 0 – 30 min Displacement from t = 0 – 30 min Average velocity for t = 0 – 30 min Instantaneous acceleration at t = 25 min

Practice Problem Write out the given info, the unknown, the equation, your working equation, and then plug and chug. Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s2, then determine the displacement of the car during the skidding process.

Free-Fall Practice Write out the given info, the unknown, the equation, your working equation, and then plug and chug. Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height.

vi a Motion V vs. T + or - + - v v t t v t v t v t v t + or - Accelerating from rest + Speeding up in positive direction - Slowing down in positive direction Speeding up in negative direction Slowing down in negative direction v v t t v t v t v t v t