1-D Kinematics Problems

Transcription

1 x (m) Name: AP Physics -D Kinemics Problems 5. Answer the following based on the elocity s. time graph straight cured a. Gie a written description of the motion. t (s) Object moes in negie direction a constant speed of 4m/s for s -4sec, object continues to moe in neg direction, but slows down to stop momentarily 4s 4-8sec, object reerses direction to moe in positie direction, speeding up to 8m/s by 8s. b. Determine the displacement from t = s to t = 4 s. x -4 = Area under -t from -4 = (-4)()+/(-4)() = - m c. Determine the displacement from t = s to t = 6 s. x -6 = Area under -t from -6 = /(-4)() +/()(4) = m d. Determine the object s accelerion t =4s. a = Slope of -t 4s = m/s e. Sketch a possible x-t graph for the motion of the object. Explain why your graph is only one of many possible graphs. 6. Answer the following based on the position s. time graph. a. Where on the graph aboe is the object moing most slowly? (How do you know?)

2 B and F because slope, which represents elocity, is approximely b. Between which points is the object speeding up? (How do you know?) between B and C, between C and D between F and G For all these interals, the magnitude of the slope (elocity) is increasing (getting steeper) c. Between which points is the object slowing down? (How do you know?) between A and B (slope is getting shallower) between D and E between E and F d. Where on the graph aboe is the object changing direction? (How do you know?) At B changes from positie to negie direction At F, changes from negie to positie direction 3. A car moes m/s and coasts up a hill with a uniform accelerion of -.6 m/s a) Wh is the displacement after 6 sec? m a.6m x t 6sec x t (6) 43.m (.6)6 b) Wh is the displacement after 9 sec? m/ s a.6m x t 9sec x t (9) 43.m (.6)9

3 c) Wh is going on? Plot -t graph of this problem to explain. Determine x from the graph. If you plot the -t graph, you see th t=, = and the slope is the accelerion which is -.6m/s. Knowing the initial point and the slope, you can find the 6 sec and 9 sec. The -t graph shows th the car goes in the positie direction (up the hill), slowing down for the first 7.5sec. At 7.5 sec, it momentarily stops. After 7.5sec, it reerses direction and speeds up (rolls down the hill speeding up). At 6 sec, car is on the way up and its displacement is 43.6 m (area under cure) At 9 sec, car is rolling back down and is the same position it was 6 sec so its displacement is the same. (area under cure from 6-9 sec is ). (m/s) t (s) 4. A speedbo starts from rest and acceleres +. m/s for 7. s. At the end of this time, the bo continues for an additional 6. s with an accelerion of +.58 m/s. Following this, the bo acceleres.49 m/s for 8. s. (a) Wh is the elocity of the bo t =. s? (b) Find the total displacement of the bo. This is a 3 segment problem! The end elocity of the first segment is the start elocity of the second, the end elocity of the second is the start elocity of the third, etc 7 sec 6 sec 8 sec a=. m/s a=.58 m/s a= -.49 m/s m a.m x t 7.s a.58m x t 6.s There aren t enough ariables in the nd and 3 rd accelerion parts to determine any unknowns. We must start in the st part and rele ariables in the st part to those in the ler parts. We know th the final elocity the end of the 7 sec is the initial elocity the start of the nd part. With th, we can find the final elocity of the nd part, which is also the initial elocity of the 3 rd part. a.49m x t 8.s

4 (.)(7) 4.7m 4.7 (.58)(6) 7.8m 7.8 (.49)(8) 5.6m x t 49.m (.)7 x t 93.7m 4.7(6) (.58)6 x total m x t 89.8m 7.8(8) (.49)8 5. Challenging: Reaction Time Problem A car is traeling m/s when the drier sees a child standing on the road. She takes.8 s to react then steps on the brakes and slows 7. m/s. How far does the car go before it stops? Reaction time a= m x t.8s x t 6m Brake a= -7m/s a 7m x t m ax x 8.6m ( 7) x x total m

5 4. m 6. On a planet th has no mosphere, a rocket 4. m tall is resting on its launch pad. Freefall accelerion on the planet is 4.45 m/s. A ball is dropped from the top of the rocket with zero initial elocity. a 4.45m y 4.m t a) how long does it take to reach the launch pad y t 4. t.53s gt ( 4.45) t b) wh is the speed of the ball just before it hits the ground? (the speed is.3 m/s, the elocity is -.3 m/s) c) wh is the speed of the ball just before it hits the ground? (the speed is.3 m/s, the elocity is -.3 m/s) f i gt ( 4.45)(.53).3m 7. You are a bungee jumping fanic and want to be the first bungee jumper on Jupiter. The length of your bungee cord is 45. m. Free fall accelerion on Jupiter is 3. m/s. Wh is the rio of your speed on Jupiter to your speed on Earth when you hae dropped 45 m? Ignore the effects of air resistance and assume th you start rest. Jupiter a 3.m y t 45m Earth a 9.8m y 45m t y f f i gy g( 45) fjupiter fearth g g Earth Jupiter ( 45) ( 45)

6 8. A stone is thrown ertically upwards with a speed of. m/s. a) How fast is it moing when it reaches a height of. m? a 9.8m y m t m.8m There are solutions: +.8 m/s when the stone is going up -.8 m/s when it is coming back down b) How long is required to reach this height? Time to get to m on the way up: gy Time to get to m on the way down c) Why are there answers to b? The rock has t 3.35s a displacement of +m twice : m aboe its starting position on the way up and m aboe its starting position on the way down. 9. A stone is thrown ertically upward with a speed of. m/s from the edge of a cliff 75. m high as shown right. m a) How much ler does it reach the bottom of the cliff? y t gt 75 t ( 9.8)() t ( 9.8) t t.73s gt gt.8 9.8t t a 9.8m y 75m y= 75m 4.9t This is a quadric equion. You can sole it by using the quadric formula to find the roots. The solution gies roots: t = 5.3 s and trock = -.88 s t 75

7 Since there is no negie time, we take the physically meaningful solution of t = 5.3 s b) Wh is its speed just before hitting? gt 9.8(5.3) 4.m c) Wh total distance did it trael? The total distance is the distance to the peak + distance back from the peak to top of cliff (same) + distance to the bottom of the cliff (75m). We first need to find y to peak of stone s flight. At the peak, =. m a 9.8m y? t gy y 7.35m ( 9.8) y ( to peak) Total distance traeled = = 89.7 m