Thursday, November 24, 2016

Soal dan Pembahasan Gerak Lurus Berubah Beraturan (GLBB)

1. While traveling along a highway a driver slows from 24 m/s to 15 m/s in 12 seconds. What is the automobile’s acceleration? (Remember that a negative value indicates a slowing down or deceleration)
Known: v1 = 24 m/s           
               v2 = 15 m/s
               t   = 12 seconds
Asked : acceleration (a)?
Answer: a = (v2 - v1) / t
               a = (15 -24) / 12
               a = - 9 / 12
               a = - 3/4 or -0.75 m/s2
So, The acceleration is -0.75 m/s2 (sign (-); indicate that the automobile is slowing down).
 
2. A parachute on a racing dragster opens and changes the speed of the car from 85 m/s to 45 m/s in a period of 4.5 seconds. What is the acceleration of the dragster?
Known: v1 = 85 m/s           
               v2 = 45 m/s
               t   = 4.5 seconds
Asked: acceleration (a)?
Answer: a = (v2 - v1) / t
               a = (45 - 85) / 4.5
               a = - 40 / 4.5
               a = -8.89 m/s2
So, The acceleration is -8.89 m/s2 (sign (-); indicate that the automobile is slowing down).

3. A car traveling at a speed of 30.0 m/s encounters an emergency and comes to a complete stop. How much time will it take for the car to stop if it decelerates at -4.0 m/s2?
Known: v1 = 30 m/s           
               v2 = 0 m/s (comes to a complete stop)
               a   = -4 m/s2
Asked: time (t)?
Answer: a = (v2 - v1) / t
               -4 = (0 - 30) / t
               -4 = - 30 / t
                t = -30/-4 
                t =  7.5 seconds
So, the time is 7.5 s.

4. If a car can go from 0 to 60 km/hr in 8.0 seconds, what would be its final speed after 5.0 seconds if its starting speed were 50 km/hr?
Known:
first case; v1 = 0 km/hr; v2 = 60 km/hr
                   t   = 8 s;
Second case;  v1 = 50 km/hr;
                        t   = 5 s;
Asked: v2?
Answer: First of all, you have to find the acceleration at first case;
before it, be carefull with the units. You have to make it in SI units.
first case; v1 = 0 km/hr
                      = 0 m/s;
                  v2 = 60 km/hr
                       = (60 x 1000)/3600 m/s
                       = 16.67 m/s
Second case;  v1 = 50 km/hr
                             = (50 x 1000)/3600 m/s
                             = 13.89;
Find the acceleration at first case;
                 a = (v2 - v1) / t
                    = (16.67 - 0) / 8
                    = - 16.67 / 8
                    = 2.08375 m/s2
Then, calculate v2(final velocity) on second case;
                 a = (v2 - v1) / t
    2.08375 = (v2 - 13.89) / 5 
2.08375 x 5= (v2 - 13.89)
                v2 = 24.30875 m/s
So, the final velocity v2 = 24.30875 m/s on second case.

5. A cart rolling down an incline for 5.0 seconds has an acceleration of 4.0 m/s2. If the cart has a beginning speed of 2.0 m/s, what is its final speed?
Known: t   = 5 seconds
               a  =  4.0 m/s2
               v1 = 2 m/s            
              
Asked: final velocity (v2)?
Answer: v2 = v1 + a . t
                    = 2 + 4 . 5
                    = 2 + 20
                    = 22 m/s
So, The final velocity is 22 m/s.

6. A car traveling at 72 km/h in 2 s is accelerating 5 m/s2. Determine the speed and the distance!
Known:  v1 = 72 km/h =20 m/s
                t   = 2 s
                a   =  5.0 m/s2
         
Asked: a). final velocity (v2)?
             b). distance (d)?

Answer: a). v2 = v1 + a . t
                         = 20 + 5 . 2
                         = 20 + 10
                         = 30 m/s
                b). d = v1 . t + 1/2 a . t2
                         = 20 . 2 + 1/2 . 5 . (2)2
                         = 40 + 10
                         = 50 m  
So, The final velocity is 30 m/s and the distance is 50 m.

7. A car moving at a speed of 72 km/h do braking so the speed regularly reduced to 18 km/h in 5 seconds. What is the distance at t = 5 s?
Known:  v1 = 72 km/h = 20 m/s
               v2 = 18 km/h = 5 m/s               
                t   = 5 s       
Asked: distance (d)?
Answer: We have to calculate the acceleration first before calculate the distance.      

                    v2 = v1 + a . t
                     5  = 20 + a . 5
                     5a= 5 - 20
                     5a= -15
                      a = -3 m/s2
Then,                      
                      d = v1 . t + 1/2 . a . t2
                         = 20 . 5 + 1/2 . -3 . (5)2
                         = 100 - 37.5
                         = 62.5 m
So, The distance is 62.5 m.

8. How long does it take for a car to change its velocity from 10 m/s to 25 m/s if the acceleration is 5 m/s2?
Known: v1 = 10 m/s           
               v2 = 25 m/s (comes to a complete stop)
               a   = 5 m/s2
Asked: time (t)?
Answer: a = (v2 - v1) / t
               5 = (25 - 10) / t
               5 = 15 / t
                t = 15/5
                t =  3 seconds
So, the time is 3 s.

9. How long will it take for an apple falling from a 29.4m-tall tree to hit the ground?
Known:  d = 29.4 m
               v1 = 0 m/s (in the beginning is at rest)
                a  = g = 10 m/s2 (gravitational acceleration)   
Asked: time (t)?
Answer:                    
                      d = v1 . t + 1/2 . a . t2
                 29.4 = 0 . t + 1/2 . 10 . (t)2
                 29.4 = 0 + 5t2
                     t2 = 29.4 / 5
                     t   = 5.881/2
                     t   = 2.42 s
So, the time is 2.42 s.

10. If a car has a constant acceleration of 4 m/s2, starting from rest, how fast is it traveling after 5 seconds?
Known: t   = 5 seconds
               a  =  4.0 m/s2
               v1 = 0 m/s (starting from rest)           
              
Asked: final velocity (v2)?
Answer: v2 = v1 + a . t
                    = 0 + 4 . 5
                    = 0 + 20
                    = 20 m/s
So, The final velocity is 20 m/s.

11. If a car has a constant acceleration of 4 m/s2, starting from rest, how far has it traveled by the time it reaches the speed of 40 m/s?
Known:  a  =  4.0 m/s2
               v1 = 0 m/s (starting from rest)           
               v2 = 40 m/s
Asked: d?
Answer: We have to find the time in the beginning, later we can calculate the distance.
               v2 = v1 + a . t
               40 = 0 + 4 . t
               40 = 4t
                 t  = 10 s
 we get t = 10 s, then
                     d = v1 . t + 1/2 . a . t2
                        = 0 . 10 + 1/2 . 4 . (10)2
                        = 200 m

So, The distance is 200 m.

12. A car is at velocity of 20 km/h. If the car traveled 120 km in 3 hours at constant acceleration, what is its final velocity?
Known:  v1 = 20 km/h (starting from rest)           
               d = 120 km
                t = 3 hours
Asked: v2?
Answer:  We have to find the acceleration in the beginning, later we can calculate the final velocity.
                     d = v1 . t + 1/2 . a . t2
                 120 = 20 . 3 + 1/2 . a . (3)2
                 120 = 60 + 4.5a
                 4.5a = 120 - 60
                 4.5a = 60
                    a   = 13.33 m/s2
Then,
               v2 = v1 + 13.33 . 3
                    = 20 + 40
                    = 60 km/h
So, the final velocity is 60 km/h.

13. How long will it take for a falling object to reach 108 m/s if its initial velocity is 10 m/s?
Known:  v1 = 10 m/s
               v2 = 108 m/s
                a  = g = 10 m/s2 (gravitational acceleration)   
Asked: time (t)?
Answer:                    
                  v2 = v1 + a . t
               108 = 10 + 10 . t
               10t = 108 - 10
               10t = 98
                    t = 9.8 s
So, the time is 9.8 s.
14. What is the final velocity of an apple if it falls from a 100m-tree?
Known:  v1 = 0 m/s (Hanging on the tree)
               d = 100 m
                a  = g = 10 m/s2 (gravitational acceleration)   
Asked: v2?
Answer:                    
                  v22 = v12 + 2. a . d
                  v22 = 02 + 2. 10 . d
                  v22 = 20 . 100
                  v2 = (2000)1/2
                   v2 = 20.(5)1/2
 So, the final velocity is 20.(5)1/2 m/s

15. What is the displacement of a car whose initial velocity is 5 m/s and then accelerated 2 m/s2 for 10 seconds?
Known:  v1 = 5 m/s           
                a = 2 m/s2
                t = 10 s
Asked: d?
Answer: 
                     d = v1 . t + 1/2 . a . t2
                        = 5 . 10 + 1/2 . 2 . (10)2
                        = 50 + 100
                        = 150 m
So, the displacement of a car is 150 m

16. What is the final velocity of a car that accelerated 10 m/s2 from rest and traveled 180m?
Known:  v1 = 0 m/s (from rest)
               d = 180 m
                a  = 10 m/s2    
Asked: v2?
Answer:                    
                  v22 = v12 + 2. a . d
                  v22 = 02 + 2. 10 . 180
                  v22 = 3600
                  v2 = (3600)1/2
                   v2 = 60 m/s
 So, the final velocity is 60 m/s

17. If a car accelerated from 5 m/s to 25 m/s in 10 seconds, how far will it travel?
Known:  v1 = 5 m/s
               v2 = 10 m/s
                t = 10 s
Asked: d?
Answer:                    
We have to find the acceleration in the beginning, later we can calculate the distance.
                  v2 = v1 + a . t
                 10 = 5 + a . 10
               10a = 10 - 5
               10a = 5
                    a = 0.5 m/s2
Then,
                     d = v1 . t + 1/2 . a . t2
                        = 5 . 10 + 1/2 . 0.5 . (10)2
                        = 50 + 25
                        = 75 m
So, the distance of car is 75 m.

18. Renata Gas is driving through town at 25.0 m/s and begins to accelerate at a constant rate of -1.0 m/s2. Eventually Rennata comes to a complete stop. Use kinematic equations to calculate the distance that Renata travels while decelerating.
Known:  v1 = 25 m/s
               v2 = 0 m/s (complete to stop)
                a = -1.0 m/s2
Asked: d?
Answer:                    
We have to find the time in the beginning, later we can calculate the distance.
                  v2 = v1 + a . t
                   0 = 25 + -1 . t
                    t = 25 s
Then,
                     d = v1 . t + 1/2 . a . t2
                        = 25 . 25 + 1/2 . -1 . (25)2
                        = 625 - 312.5
                        = 312.5 m
So, the distance of car is 312.5 m.

19. An object that is not moving initially given force F so that the speed is 10 m / s and the distance of    15 meters. Calculate the length of time the object is given force F!
Known:  v1 = 0 m/s (from rest)
               v2 = 10 m/s
                 d = 15 m    
Asked: t?
Answer:  We calculate the acceleration at first,                  
                  v22 = v12 + 2. a . d
                  102 = 02 + 2. a . 15
                  100 = 30a
                  a = 100/30
                  a = 3.33 m/s2
Then,
                  v2 = v1 + a . t
                  10 = 0 + 3.33 . t
                    t = 3 s

 So, the time is 3 s.




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