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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