# Lecture 04 Kinematics Dynamics l Kinematics Equations constant

- Slides: 9

Lecture 04: Kinematics + Dynamics l Kinematics Equations èconstant acceleration l Dynamics èNewton’s Second Law èNon-zero acceleration

Equations for Constant Acceleration lx lv = x 0 + v 0 t + ½ at 2 = v 0 + at l v 2 = v 02 + 2 a(x-x 0) Ø x is final position Ø xo is initial position Ø v is final velocity Ø v 0 is initial velocity Ø a is acceleration Ø t is time

Kinematics Example l Ø A car is traveling 30 m/s and applies its breaks to stop. Assuming constant acceleration of -6 m/s 2, how long does it take for the car to stop, and how far does it travel before stopping? Begin by using the second equation to find the time: Ø v = v 0 + a t l x = x 0 + v 0 t + ½ at 2 l v = v 0 + at l v 2 = v 02 + 2 a(x-x 0) Then use the first equation to find the distance: (0 m/s) = (30 m/s) + (-6 m/s 2) t t=5 s x = x 0 + v 0 t + ½ a t 2 x = (0 m) + (30 m/s)(5 s) + ½ (-6 m/s 2) (5 s)2 x = 75 m

Dynamics: F = ma l We have already dealt with situations where a = 0. l But when the net force is not zero, there IS an acceleration!

Dynamics Example l A tractor is pulling a trailer with a constant acceleration. If the forward acceleration is 1. 5 m/s 2, Calculate the force on the trailer (m = 400 kg) due to the tractor. (Consider just the trailer. ) FBD: x-direction: FN FT = ma FT Fg y F = ma FT = 600 N y-direction: F = ma FN - F g = 0 x FN = 3920 N FT = 600 N

Summary of Concepts • Constant Acceleration Ø x = x 0 + v 0 t + ½ at 2 Ø v = v 0 + at Ø v 2 = v 02 + 2 a(x-x 0) • F = m a – Draw Free Body Diagram – Write down equations – Solve

Kinematics Example l A car moving at 15 m/s is traveling toward an intersection and sees the light turn yellow. The car accelerates at 4 m/s 2 until it gets to the intersection 18 m away. How long does it take the car to get to the intersection? (And assuming the light is yellow for 1 s, does the car make it before the light turns red? ) Note: even after accelerating, the car is still traveling safely under the speed limit… l There are two ways to solve this…I will use one method and you will use the other!

Kinematics Example l A car moving at 15 m/s is traveling toward an intersection and sees the light turn yellow. The car accelerates at 4 m/s 2 until it gets to the intersection 18 m away. How long does it take the car to get to the intersection? (And assuming the light is yellow for 1 s, does the car make it before the light turns red? ) Note: even after accelerating, the car is still traveling safely under the speed limit… l I will do this in two steps: è First, using v 2 = v 02 + 2 a (x-x 0), we can solve for the final velocity. è Second, using v = v 0 + a t, we can solve for the time.

Kinematics Example l A car moving at 15 m/s is traveling toward an intersection and sees the light turn yellow. The car accelerates at 4 m/s 2 until it gets to the intersection 18 m away. How long does it take the car to get to the intersection? (And assuming the light is yellow for 1 s, does the car make it before the light turns red? ) Note: even after accelerating, the car is still traveling safely under the speed limit… v 2 = v 02 + 2 a (x-x 0) v 2 = (15 m/s)2 + 2 (4 m/s 2) (18 m – 0 m) v = 19. 2 m/s v = v 0 + a t (19. 2 m/s) = (15 m/s) + (4 m/s 2) t t = 1. 05 s The car does not make it before the light turns red… It should have just stopped!

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