The Magic Chart Honors Physics

Transcription

1 The Magic Chart Honors Physics Magic Chart Equations v f = v i + a t x = v i t + 1/2 a t 2 x = ½ (v i + v f ) t v 2 f = v 2 i + 2a x x = v f t - 1/2 a t 2 x Who Cares Quantity v f a t V i THE WHO CARES QUANTITY tells you which equation to use. For example, suppose you know a, x, and v i but not v or t. If you are looking for t, then you don't care about v, so use the equation whose WHO CARES quantity is v. If, however, you need to find v, use the equation whose WHO CARES quantity is t. Now try these on your own: 1. You throw an object upwards from the ground with a velocity of 20m/s and it is subject to a downward acceleration of 9.8 m/s 2. How high does it go? v i=20 m/s a=-9.8 m/s 2 vf 2 = vi 2 + 2a x (0 m/s) 2 = (20 m/s) (- 9.8 m/s 2 ) x x=20.4 meters 2. You throw an object upwards from the ground with a velocity of 20m/s and it is subject to a downward acceleration of -9.8 m/ s 2. How much later does it momentarily coming to a stop? v i=20 m/s a=-9.8 m/s 2 vf = vi + a t =(20 m/s) m/s 2 t t=2.04 s 3. You throw an object upwards from the ground with a velocity of 20m/s and it is subject to an downward acceleration of 9.8 m/ s 2. How high is it after 2.0s? v i=20 m/s t=2 s vf 2 = vi 2 + 2a x =(20 m/s) (-9.8 m/s 2 )x =20.4 m 1 Page

2 What about problems that act along a different axis? How does the context of the problem change Example 2 A car is at rest when it experiences an acceleration of 2.0m/s 2 towards the north for 5.0s. How far will it travel during the time it accelerates? Take a second to write down the facts in the problem and then determine which equation you would use. Since we re not told where the car starts, let s just define its initial position as the origin for this problem, zero. In that case, the distance it travels will just be its position, x, at the end of the problem. Then, x0 = 0 x =? vo = 0 t= 5.0s a = 2.0m/s 2 x = vot + ½at 2 Example 3 An object accelerates from rest. How long will it take for it to travel 40m if its acceleration is 4m/s 2? a= 4 m/s 2 x=40 m x = vi t + 1/2 a t 2 40 m = ½ (4 m/s 2 )t =20 s Example 4 A plane must reach a speed of 36m/s in order to take off and its maximum acceleration is 3.0m/s 2. How long a runway does it require? v f=36 m/s a= 4 m/s 2 vf 2 = vi 2 + 2a x (36 m/s) 2 =2(4m/s 2 )x =162 m 2 Page

3 Problem Solving with Kinematics Equation 2 Class Work 1. An object accelerates from rest with a constant acceleration of 2 m/s 2. How far will it have moved after 9s? a= 2 m/s 2 t=9 s x = vi t + 1/2 a t 2 = ½ (2 m/s 2 )(9 s) 2 =81 m 2. An object, moving with constant acceleration and zero initial velocity, travels 48 m in 5.2 s. What is the magnitude of its acceleration? a= t=5.2 s x=48 m x = vi t + 1/2 a t 2 48 m = ½ a (5.2 s) 2 =3.55 m/s/s 3. An object, initially at rest, is subject to an acceleration of 34 m/s 2. How long will it take for that object to travel 3400 m? a= 34 m/s 2 x=3400 m x = vi t + 1/2 a t m = ½ (34 m/s 2 )t =14.1 s 4. An object is shot upwards, from the ground, with an initial velocity of 120 m/s. How high will it be after 4.0s? v i=120 m/s t=4 s x = vi t + 1/2 a t 2 =(120 m/s) 2 (4 s)+1/2(-9.8 m/s 2 )(4 s) 2 =401.6 m 5. An object, initially at rest, moves 250 m in 17 s. What is its acceleration? a= t=17 s x=250 m x = vi t + 1/2 a t m = ½ a (17 s) 2 =1.73 m/s/s 3 Page

4 6. An object is shot upwards, from the ground, with an initial velocity of 40 m/s. How high will it be after 2.0s? v i=40 m/s t=2 s x = vi t + 1/2 a t 2 =(40 m/s) 2 (2 s)+1/2(-9.8 m/s 2 )(2 s) 2 =60.4 m 7. An object is dropped from a 42 m tall building. How long does it take to reach the ground? x=42 m x = vi t + 1/2 a t 2 42 m = ½ (-9.8 m/s 2 )t =2.9 s Homework 8. An object is moving with an initial velocity of 23 m/s. It is then subject to a constant acceleration of 3.5 m/s 2 for 12 s. How far will it have traveled during the time of its acceleration? v i=23 m/s t=12 s a= 3.5 m/s 2 x = vi t + 1/2 a t 2 =(23 m/s) 2 (12 s)+1/2(3.5 m/s 2 )(2 s) 2 =528 m 9. An object is at rest when it undergoes a constant acceleration of 13 m/s2 for 5.0 seconds. How far will it have traveled during this time? t=5 s a= 13 m/s 2 x = vi t + 1/2 a t 2 =1/2(13m/s 2 )(5 s) 2 =162.5 m 10. An object is dropped from the top of a building and strikes the ground 2.0s later. How tall is the building? t=2 s x = vi t + 1/2 a t 2 = ½ (-9.8 m/s 2 )(2 s) 2 =19.6 m 4 Page

5 Problem Solving with Kinematics Equation 3 Class Work 11. An object accelerates from rest with a constant acceleration of 7.5 m/s 2. How fast will it be traveling after it goes 21 m? a= 7.5 m/s 2 v f=? x=21 m vf 2 = vi 2 + 2a x = 2(7.5 m/s 2 )(21 m) =17.74 m/s 12. An object experiences an acceleration of 9.8 m/s 2 over a distance of 210 m. After that acceleration it has a velocity of 380 m/s. What was its initial velocity before being accelerated? v i=? a= 9.8 m/s 2 v f=380 m/s x=210 m vf 2 = vi 2 + 2a x (380 m/s) 2 =v i2 + 2(9.8 m/s 2 )(210 m) =5.91 m/s 13. An object accelerates from rest to 24 m/s over a distance of 56 m. What acceleration did it experience? a=? v f=24 m/s x=56 m vf 2 = vi 2 + 2a x (24 m/s) 2 =2(a)(56 m) =5.14 m/s A car which is traveling at a velocity of 15 m/s undergoes an acceleration of 6.5 m/s 2 over a distance of 340 m. How fast is it going after that acceleration? v i=15 m/s a= 6.5 m/s 2 v f=? x=340 m vf 2 = vi 2 + 2a x = (15 m/s) 2 + 2(6.5 m/s 2 )(340 m) =68.15 m/s 15. A car slams on its brakes creating an acceleration of -3.2 m/s 2. It comes to rest after traveling a distance of 210 m. What was its velocity before it began to accelerate? v i=? a= -3.2 m/s 2 x=210 m vf 2 = vi 2 + 2a x 0 =v i2 + 2(-3.2 m/s 2 )(210 m) =36.6 m/s 5 Page

6 Homework 16. An object experiences an acceleration of -6.8 m/s 2. As a result, it accelerates from 54 m/s to a complete stop. How much distance did it travel during that acceleration? v i=54 a= -6.8 m/s 2 m vf 2 = vi 2 + 2a x (0 m/s) 2 =(54 m/s) 2 + 2(- 6.8 m/s 2 )x = m 17. An object is dropped from a 32 m tall building. How fast will it be going when it strikes the ground? v f=? m/s x=32 m vf 2 = vi 2 + 2a x = 2(-9.8 m/s 2 )(32 m) =25.04 m/s 18. An object is dropped from a building and strikes the ground with a speed of 31m/s. How tall is the building? v f=31 m/s m vf 2 = vi 2 + 2a x (31 m/s) 2 = 2(-9.8 m/s 2 )x =49.03 m 19. A hopper jumps straight up to a height of 1.3 m. With what velocity did it leave the floor? v i=? m/s x=1.3 m vf 2 = vi 2 + 2a x = vi 2 +2(-9.8 m/s 2 )(1.3 m) =5.04 m/s 20. A hopper jumps straight up to a height of 0.45 m. With what velocity will it return to the table? v i=? m/s x=0.45 m vf 2 = vi 2 + 2a x = vi 2 +2(-9.8 m/s 2 )(0.45 m) =2.96 m/s 6 Page