8.10 Code Practice: Question 2

Earlier in Lesson 6, four kinematic equations were introduced and discussed. A useful problem-solving strategy was presented for use with these equations and 2 examples were given that illustrated the utilize of the strategy. And then, the application of the kinematic equations and the trouble-solving strategy to free-fall move was discussed and illustrated. In this part of Lesson 6, several sample issues volition exist presented. These problems allow any pupil of physics to test their agreement of the use of the four kinematic equations to solve problems involving the i-dimensional motion of objects. Y'all are encouraged to read each trouble and do the use of the strategy in the solution of the problem. So click the button to check the answer or apply the link to view the solution.

Bank check Your Understanding

1. An plane accelerates downward a rails at three.20 thou/southward² for 32.eight due south until is finally lifts off the basis. Determine the distance traveled earlier takeoff.
2. A auto starts from rest and accelerates uniformly over a time of 5.21 seconds for a distance of 110 one thousand. Determine the acceleration of the car.
3. Upton Chuck is riding the Behemothic Drop at Great America. If Upton free falls for 2.threescore seconds, what will be his final velocity and how far volition he fall?
4. A race automobile accelerates uniformly from 18.5 1000/south to 46.1 m/s in two.47 seconds. Decide the acceleration of the car and the distance traveled.
5. A plume is dropped on the moon from a top of 1.40 meters. The dispatch of gravity on the moon is one.67 m/s². Make up one's mind the time for the feather to autumn to the surface of the moon.
6. Rocket-powered sleds are used to examination the man response to acceleration. If a rocket-powered sled is accelerated to a speed of 444 one thousand/s in i.83 seconds, then what is the acceleration and what is the altitude that the sled travels?
7. A bike accelerates uniformly from residual to a speed of vii.10 1000/s over a distance of 35.4 m. Determine the acceleration of the bike.
8. An engineer is designing the runway for an airport. Of the planes that volition use the airport, the lowest acceleration rate is likely to be three one thousand/sⁱⁱ. The takeoff speed for this aeroplane volition be 65 m/s. Assuming this minimum acceleration, what is the minimum allowed length for the rails?
9. A car traveling at 22.4 m/s skids to a stop in 2.55 s. Determine the skidding distance of the auto (assume uniform dispatch).
10. A kangaroo is capable of jumping to a height of 2.62 m. Determine the takeoff speed of the kangaroo.
11. If Michael Hashemite kingdom of jordan has a vertical leap of 1.29 m, then what is his takeoff speed and his hang fourth dimension (full time to motion upward to the pinnacle and so return to the ground)?
12. A bullet leaves a rifle with a muzzle velocity of 521 m/s. While accelerating through the butt of the rifle, the bullet moves a distance of 0.840 thousand. Determine the acceleration of the bullet (assume a uniform acceleration).
13. A baseball is popped straight upwardly into the air and has a hang-time of six.25 s. Decide the height to which the ball rises before it reaches its meridian. (Hint: the time to rise to the summit is 1-half the total hang-time.)
14. The observation deck of alpine skyscraper 370 m above the street. Determine the time required for a penny to free autumn from the deck to the street below.
15. A bullet is moving at a speed of 367 k/southward when information technology embeds into a lump of moist clay. The bullet penetrates for a distance of 0.0621 1000. Decide the acceleration of the bullet while moving into the clay. (Assume a uniform acceleration.)
16. A stone is dropped into a deep well and is heard to hit the water 3.41 s after being dropped. Determine the depth of the well.
17. It was one time recorded that a Jaguar left skid marks that were 290 m in length. Assuming that the Jaguar skidded to a stop with a constant acceleration of -3.90 k/sⁱⁱ, determine the speed of the Jaguar before information technology began to skid.
18. A plane has a takeoff speed of 88.3 m/s and requires 1365 thou to accomplish that speed. Determine the dispatch of the aeroplane and the time required to achieve this speed.
19. A dragster accelerates to a speed of 112 m/s over a distance of 398 yard. Determine the acceleration (assume compatible) of the dragster.
20. With what speed in miles/hr (1 grand/s = 2.23 mi/hour) must an object be thrown to reach a acme of 91.v one thousand (equivalent to one football field)? Presume negligible air resistance.

Solutions to Higher up Problems

1. Given: a = +iii.two chiliad/southward2 t = 32.8 s vi = 0 m/s

   Observe: d = ??

   d = v_i*t + 0.5*a*t²

   d = (0 m/s)*(32.8 southward)+ 0.5*(three.20 chiliad/s^two)*(32.8 south)²

   d = 1720 m

   Return to Trouble 1

2. Given: d = 110 m t = 5.21 southward vi = 0 thou/south

   Find: a = ??

   d = v_i*t + 0.5*a*t²

   110 m = (0 thou/s)*(5.21 due south)+ 0.5*(a)*(v.21 southward)²

   110 thou = (13.57 s²)*a

   a = (110 m)/(xiii.57 s^two)

   a = 8.10 m/ s²

   Render to Problem two

3. Given: a = -9.8 m t = 2.half dozen south vi = 0 g/south

   Find: d = ?? fivef = ??

   d = v_i*t + 0.5*a*t^two

   d = (0 m/south)*(2.lx s)+ 0.v*(-9.8 m/s^two)*(2.60 south)²

   d = -33.1 one thousand (- indicates direction)

   v_f = v_i + a*t

   v_f = 0 + (-9.8 m/southward^two)*(two.60 due south)

   v_f = -25.5 thousand/s (- indicates management)

   Render to Trouble 3

4. Given: 5i = 18.5 one thousand/s vf = 46.ane 1000/southward t = 2.47 south

   Detect: d = ?? a = ??

   a = (Delta v)/t

   a = (46.1 yard/s - 18.five k/s)/(2.47 s)

   a = 11.2 yard/south²

   d = v_i*t + 0.5*a*t²

   d = (18.v m/s)*(2.47 s)+ 0.5*(11.2 m/s^two)*(ii.47 s)²

   d = 45.vii yard + 34.one 1000

   d = 79.eight chiliad

   (Note: the d tin also be calculated using the equation 5_f ⁱⁱ = v_i ² + 2*a*d)

   Render to Problem 4

5. Given: vi = 0 thousand/due south d = -1.40 m a = -1.67 m/south2

   Find: t = ??

   d = five_i*t + 0.5*a*t²

   -1.40 m = (0 m/s)*(t)+ 0.5*(-1.67 m/s²)*(t)ⁱⁱ

   -ane.40 m = 0+ (-0.835 grand/s²)*(t)^two

   (-one.twoscore m)/(-0.835 m/s²) = t²

   i.68 s^two = t²

   t = 1.29 s

   Return to Problem v

6. Given: 5i = 0 m/s fivef = 444 thousand/s t = 1.83 s

   Find: a = ?? d = ??

   a = (Delta v)/t

   a = (444 m/due south - 0 thou/southward)/(i.83 southward)

   a = 243 m/due south²

   d = v_i*t + 0.v*a*t²

   d = (0 yard/s)*(1.83 s)+ 0.five*(243 m/s²)*(one.83 s)^two

   d = 0 m + 406 m

   d = 406 one thousand

   (Note: the d can likewise be calculated using the equation v_f ² = v_i ^two + 2*a*d)

   Return to Trouble vi

   
   

7. Given: fivei = 0 m/s vf = 7.ten m/s d = 35.four yard

   Observe: a = ??

   5_f ⁱⁱ = v_i ² + 2*a*d

   (7.10 thousand/due south)² = (0 m/southward)² + 2*(a)*(35.iv m)

   l.iv m²/s² = (0 m/s)² + (lxx.viii m)*a

   (50.four grand²/south²)/(70.8 thou) = a

   a = 0.712 one thousand/s²

   Return to Trouble 7

8. Given: vi = 0 m/s vf = 65 g/due south a = three g/s2

   Observe: d = ??

   v_f ² = v_i ² + 2*a*d

   (65 m/southward)^two = (0 m/s)² + 2*(3 m/sⁱⁱ)*d

   4225 m²/south^two = (0 k/s)² + (6 m/s²)*d

   (4225 m^two/s²)/(half dozen grand/southward^two) = d

   d = 704 m

   Return to Problem 8

9. Given: fivei = 22.4 thou/s vf = 0 m/s t = 2.55 southward

   Notice: d = ??

   d = (v_i + v_f)/2 *t

   d = (22.four g/s + 0 m/due south)/two *2.55 s

   d = (11.two grand/south)*2.55 south

   d = 28.6 1000

   Return to Trouble 9

10. Given: a = -ix.8 g/stwo vf = 0 m/southward d = 2.62 m

    Notice: vi = ??

    v_f ² = five_i ² + 2*a*d

    (0 yard/s)² = 5_i ² + two*(-9.eight m/s²)*(two.62 m)

    0 m²/due south² = v_i ² - 51.35 mⁱⁱ/s^two

    51.35 thousand²/s² = v_i ²

    v_i = 7.17 m/s

    Return to Problem 10

11. Given: a = -nine.8 grand/s2 vf = 0 m/s d = 1.29 m

    Find: vi = ?? t = ??

    v_f ² = five_i ⁱⁱ + 2*a*d

    (0 m/s)^two = five_i ⁱⁱ + 2*(-9.8 g/s^two)*(one.29 m)

    0 m²/due south² = v_i ² - 25.28 thouⁱⁱ/s^two

    25.28 one thousand²/s² = 5_i ^two

    five_i = 5.03 m/due south

    To notice hang time, discover the fourth dimension to the peak and and so double information technology.

    v_f = v_i + a*t

    0 m/south = 5.03 m/s + (-ix.8 m/s²)*t_upwardly

    -5.03 thou/s = (-9.8 m/sⁱⁱ)*t_up

    (-5.03 m/s)/(-ix.8 grand/due south²) = t_upwardly

    t_up = 0.513 s

    hang fourth dimension = 1.03 south

    Return to Trouble eleven

12. Given: vi = 0 m/s 5f = 521 m/s d = 0.840 thousand

    Observe: a = ??

    v_f ⁱⁱ = v_i ² + 2*a*d

    (521 k/s)ⁱⁱ = (0 m/due south)² + 2*(a)*(0.840 chiliad)

    271441 m²/s² = (0 one thousand/s)² + (1.68 yard)*a

    (271441 m²/southward²)/(1.68 g) = a

    a = i.62*10^v m /s²

    Return to Problem 12

13. Given: a = -9.viii 1000/southward2 fivef = 0 one thousand/s t = 3.13 s

    Detect: d = ??

    1. (Annotation: the time required to move to the tiptop of the trajectory is one-half the total hang fourth dimension - iii.125 southward.)

    Beginning apply:  v_f = 5_i + a*t

    0 k/s = v_i + (-9.viiim/south² )*(3.thirteen s)

    0 m/due south = five_i - xxx.7 m/s

    five_i = thirty.7 m/s  (30.674 g/due south)

    Now utilize:  five_f ⁱⁱ = 5_i ² + 2*a*d

    (0 m/due south)² = (thirty.vii m/s)^two + 2*(-ix.8thousand/s² )*(d)

    0 thousand²/s^two = (940 thousand ² /s ² ) + (-19.half dozenm/sⁱⁱ )*d

    -940m ^two /s ² = (-xix.half dozenm/south² )*d

    (-940m ^two /south ²)/(-19.vim/sⁱⁱ ) = d

    d = 48.0 m

    Return to Trouble xiii

14. Given: fivei = 0 m/s d = -370 yard a = -9.viii thousand/stwo

    Find: t = ??

    d = 5_i*t + 0.5*a*t²

    -370 thou = (0 m/s)*(t)+ 0.5*(-9.8 m/southward²)*(t)²

    -370 thousand = 0+ (-4.9 chiliad/s²)*(t)²

    (-370 m)/(-4.9 m/southward²) = t²

    75.5 s^two = t^two

    t = eight.69 south

    Render to Trouble fourteen

15. Given: vi = 367 m/s vf = 0 m/s d = 0.0621 thou

    Find: a = ??

    v_f ² = v_i ² + two*a*d

    (0 m/south)^two = (367 m/s)² + 2*(a)*(0.0621 grand)

    0 k²/southwardⁱⁱ = (134689 m^two/s²) + (0.1242 chiliad)*a

    -134689 g²/s² = (0.1242 m)*a

    (-134689 thousand²/s²)/(0.1242 m) = a

    a = -1.08*10^vi m /s²

    (The - sign indicates that the bullet slowed down.)

    Render to Problem 15

16. Given: a = -9.8 m/s2 t = iii.41 due south vi = 0 chiliad/s

    Notice: d = ??

    d = v_i*t + 0.5*a*t²

    d = (0 m/s)*(iii.41 southward)+ 0.5*(-9.8 m/s²)*(3.41 s)²

    d = 0 yard+ 0.5*(-nine.viii k/s²)*(eleven.63 sⁱⁱ)

    d = -57.0 m

    (NOTE: the - sign indicates direction)

    Return to Problem xvi

17. Given: a = -three.90 chiliad/s2 vf = 0 m/southward d = 290 m

    Find: vi = ??

    five_f ² = five_i ^two + ii*a*d

    (0 m/southward)² = v_i ² + two*(-iii.90 m/s² )*(290 m)

    0 grand²/s² = v_i ^two - 2262 m²/sⁱⁱ

    2262 gⁱⁱ/s² = 5_i ²

    five_i = 47.half dozen k /south

    Return to Problem 17

18. Given: vi = 0 yard/south vf = 88.3 k/due south d = 1365 grand

    Find: a = ?? t = ??

    v_f ² = 5_i ² + 2*a*d

    (88.3 m/south)ⁱⁱ = (0 m/s)ⁱⁱ + ii*(a)*(1365 chiliad)

    7797 one thousand²/s² = (0 grand²/s²) + (2730 one thousand)*a

    7797 m²/s² = (2730 1000)*a

    (7797 gⁱⁱ/sⁱⁱ)/(2730 k) = a

    a = 2.86 yard/southⁱⁱ

    five_f = v_i + a*t

    88.3 m/due south = 0 g/s + (ii.86 1000/due south²)*t

    (88.3 m/southward)/(2.86 chiliad/s²) = t

    t = 30. 8 s

    Return to Trouble 18

19. Given: vi = 0 m/s vf = 112 thousand/s d = 398 k

    Notice: a = ??

    v_f ² = v_i ⁱⁱ + ii*a*d

    (112 m/southward)² = (0 m/southward)² + 2*(a)*(398 thousand)

    12544 chiliad^two/sⁱⁱ = 0 m²/s^two + (796 grand)*a

    12544 1000²/south² = (796 m)*a

    (12544 m²/s^two)/(796 m) = a

    a = 15.8 m/s²

    Return to Problem nineteen

20. Given: a = -9.8 thou/sii vf = 0 m/due south d = 91.5 m

    Find: fivei = ?? t = ??

    Beginning, discover speed in units of m/south:

    v_f ² = 5_i ² + 2*a*d

    (0 chiliad/due south)² = 5_i ^two + 2*(-nine.8 m/s²)*(91.five m)

    0 thou^two/s² = v_i ^two - 1793 one thousand²/sⁱⁱ

    1793 grand²/sⁱⁱ = v_i ²

    v_i = 42.3 m/s

    At present catechumen from m/s to mi/hr:

    5_i = 42.3 m/s * (two.23 mi/hr)/(one k/s)

    5_i = 94.4 mi/60 minutes

    Return to Problem xx

8.10 Code Practice: Question 2,

Source: https://www.physicsclassroom.com/Class/1DKin/U1L6d.cfm

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