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ACCELERATION

ACCELERATION. CH2 SEC 2. ACCELERATION MEASURES THE RATE OF CHANGE IN VELOCITY. AVERAGE ACCELERATION EQUATION. EXAMPLE PROBLEM. A SHUTTLE BUS SLOWS TO A STOP WITH AN AVG ACCELERATION OF -1.8 m/s 2 . How long does it take the bus to slow from 9.0 m/s to 0.0 m/s?

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ACCELERATION

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  1. ACCELERATION CH2 SEC 2

  2. ACCELERATION MEASURES THE RATE OF CHANGE IN VELOCITY • AVERAGE ACCELERATION EQUATION

  3. EXAMPLE PROBLEM • A SHUTTLE BUS SLOWS TO A STOP WITH AN AVG ACCELERATION OF -1.8 m/s2. How long does it take the bus to slow from 9.0 m/sto 0.0 m/s? • Given: vi= 90 m/s vf= 0 m/s aavg= -1.8 m/s2 • Unknown: ∆t=? • Solve: Rearrange the avg. acceleration equation to solve for the time interval 1st change this: to: 2rd solve: Answer:

  4. ACCELERATION HAS DIRECTION AND MAGNITUDE • Imagine a train is moving to the right (East), so that the displacement and the velocity are positive. • Velocity increases in magnitude as the train picks up speed. (so, when ∆v is positive, the acceleration is positive)=A • The train with no stops will travel for awhile at a constant velocity; because the velocity is not changing, ∆v= 0m/s.=B (when velocity is constant, the acceleration is equal to zero) • If the train still traveling in a positive direction, slows down, the velocity is positive, but the acceleration is negative.=C

  5. RECOGNIZING MOTION ON A GRAPH • A= speed or velocity increasing • B= constant velocity • C= speed or velocity decreasing

  6. VELOCITY AND ACCELERATION TABLE

  7. MOTION WITH CONSTANT ACCELERATION • A ball moving in a straight line with constant acceleration. (image was captured 10 times in 1 second, so the time interval between images equals 1/10 of a second) • As the ball’s velocity increases, the ball travels a greater distance during each time interval. • Because the acceleration is constant, the velocity increases by the same amount during each time interval. • So, the distance that the ball travels in each time interval is equal to the distance it traveled in the previous time interval, plus a constant distance.

  8. The relationships between displacement, velocity, and constant acceleration are expressed by equations that apply to any object moving with constant acceleration.

  9. Displacement depends on acceleration, initial velocity, and time • We know that the average velocity equals displacement divided by the time interval. • For an object moving with constant acceleration, the avg velocity is equal to the average of the initial velocity and the final velocity. • To find an expression for the displacement in terms of the initial and final velocity, we can set the expressions for average velocity equal to each other. • This equation can be used to find the displacement of any object moving with constant acceleration.

  10. Displacement with Constant Uniform Acceleration Equation Displacement= ½ (initial velocity + final velocity)(time intervial)

  11. Example Question • A racing car reaches a speed of 42 m/s. It then begins a uniform negative acceleration, using its parachute and braking system, and comes to a rest 5.5 seconds later. Find how far the car moves while stopping. Given: vi= 42 m/s vf= 0 m/s ∆t= 5.5 s Unknown: ∆x=? Solve: Use the equation for displacement

  12. Final VelocityDepends on initial velocity, acceleration, and time • What if final velocity is not known, but we still want to calculate the displacement? (we can if we known the initial velocity, the uniform acceleration, and the elapsed time) • By rearranging the equation for acceleration, we can find a value for finial velocity. • Adding the initial velocity to both sides of the equation, we get an equation for the final velocity.

  13. Velocity with Constant Uniform Acceleration Final velocity= initial velocity + (acceleration x time intervial) You can use this equation to find the final velocity of an object moving with uniform acceleration after it has accelerated at a constant rate for any time interval, whether the time interval is a minute or half an hour.

  14. If you want to know the displacement of an object moving with uniform acceleration over some certain time interval, you can obtain another useful expression for displacement by substituting the expression for vf into the expression for ∆x. • This equation is useful not only for finding displacement of an object with uniform acceleration, but also for finding the displacement required for an object to reach a certain speed or to come to a stop.

  15. Example question • A plane started at rest at one end of a runway undergoes a uniform acceleration of 4.8 m/s2 for 15 s before takeoff. What is its speed at takeoff? How long must the runway be for the plane to be able to take off? Given: vi= 0 m/s a= 4.8 m/s ∆t= 15 s Unknown: vf=? ∆x=? Solve: Use the equation for the velocity of a uniformly accelerated object. Part 2:Use the equation for the displacement

  16. Time Can Also Be Found • Method involves rearranging one equation to solve for ∆t and substituting that expression in another equation. • Start with the equation for displacement. • We can sub this expression into the equation for final velocity.

  17. Final Velocity After Any Displacement • When using this equation, you must take the square root of the right side of the equation to find the final velocity.

  18. Example Problem • A person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500 m/s2. What is the velocity of the stroller after it has traveled 4.75m? Given: vi= 0m/s a= 0.500 m/s2 ∆x= 4.75 m Unknown: vf=? Solve: Use final velocity equation. Substitute the values into the equation: Remember to take the square root in the final step.

  19. EQUATIONS FOR UNIFORMLY ACCELERATED STRAIGHT-LINE MOTION

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