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Physics

Kinetics

1. You go on vacation by car on the highway for 3 hours at a speed of 110 km·h-1. Then you stop for 30 minutes. You continue with a two-hour drive at a constant speed of 90 km·h-1 to the destination. Determine the average travel speed.

Solution:

Analysis:

physics-kinematics-1

The average travel speed was v = 92.73 km·h-1.

2.The highway section is 25 km long. The maximum allowed speed is 110 km·h-1. The driver covered this section in 12 minutes. Did they exceed the maximum allowed speed on the highway?

Solution:

Analysis:

s = 25 km = 2500 m;   t = 12 min = 720 s.

physics-kinematics-2

The driver exceeded the allowed speed by 15 km·h-1.

3.A convoy of military vehicles is 2 km long. It moves at a constant speed of 30 km·h-1. The distance from the front of the convoy to the last vehicle was covered by a courier at an average speed of 50 km·h-1. Back, at a speed of 60 km·h-1. How much time did the courier need and what distance did they cover?

Solution:

Analysis:
d=2 km, v1=30 km.h−1, v2=50 km.h−1, v3=60 km.h−1, d1=d2=2 kmd = 2\,\text{km},\ v_1 = 30\,\text{km·h}^{-1},\ v_2 = 50\,\text{km·h}^{-1},\ v_3 = 60\,\text{km·h}^{-1},\ d_1 = d_2 = 2\,\text{km}

In the opposite direction of the convoy:

d1=(v2+v1)t1⇒t1=2 km50 km.h−1+30 km.h−1=0.025 h=1.5 mind_1 = (v_2 + v_1)t_1 \Rightarrow t_1 = \frac{2\,\text{km}}{50\,\text{km·h}^{-1} + 30\,\text{km·h}^{-1}} = 0.025\,\text{h} = 1.5\,\text{min}

In the same direction as the convoy:

d2=(v3−v1)t2⇒t2=2 km60 km.h−1−30 km.h−1=115 h=4 mind_2 = (v_3 - v_1)t_2 \Rightarrow t_2 = \frac{2\,\text{km}}{60\,\text{km·h}^{-1} - 30\,\text{km·h}^{-1}} = \frac{1}{15}\,\text{h} = 4\,\text{min}

Total time:
                                                      t=t1+t2=1.5 min+4 min=5.5 mint = t_1 + t_2 = 1.5\,\text{min} + 4\,\text{min} = 5.5\,\text{min}

Total distance:s=d1+d2=(v2+v1)t1+(v3−v1)t2=(50+30)⋅0.025+(60−30)⋅115=2 km+2 km=4 kms = d_1 + d_2 = (v_2 + v_1)t_1 + (v_3 - v_1)t_2 = (50 + 30)·0.025 + (60 - 30)·\frac{1}{15} = 2\,\text{km} + 2\,\text{km} = 4\,\text{


4. A freely falling body has a speed of 3.0 m·s-1 at point A and a speed of 7.0 m·s-1 at a lower point B. Determine the time to cover distance AB. What is the distance between points A and B? With what speed will the body hit the ground if its motion from point B to the ground lasts another 2 s?

Solution:

Analysis:

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5.A swimmer whose speed relative to the water is 0.85 m·s-1 swims in a river where the water flows at 0.40 m·s-1. Determine the time to swim from point A to B, which are 90 m apart, if the swimmer swims:

  • a) downstream
  • b) upstream
  • c) perpendicular to the current (the resultant speed is perpendicular to the current speed).

Solution:

Analysis:

v1 = 0.85 m.s-1, v2 = 0.4 m.s-1, s = 90m

  • downstream
kinetics-r-1
  • upstream
kinetics-r-2 
  • perpendicular to the current 
kinetics-r-3

The swimmer reaches B from A downstream in t1 = 72 s, upstream in t2 = 200 s, and perpendicular to the current in t3 = 120s.

6. A motorboat traveling on a river covered a distance of 120 m when going downstream in 14 s, and when going upstream in 24 s. Determine the boat’s speed v1 relative to the water and the river current speed v2.

Solution:

Analysis:

v1 – boat speed relative to the water, v2 – river current speed

t1 = 14 s, t2 = 24 s

physics-kinematics-6 

The boat speed relative to the water is 6.78 m·s-1, the river current speed is 1.78 m·s-1.

7.A motorist covered the first third of the distance at a constant speed v1, and the remaining two thirds at v2 = 72 km·h-1. The average speed v was 36 km·h-1. Calculate v1!

Solution:

Analysis:

v1 = ?, v2 = 72 km·h-1 = 20 m·s-1, v = 36 km·h-1 = 10 m·s-1

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The first third of the distance was covered at v1 = 5 m·s-1.

8.A car traveling at 90 km·h-1 saw a stationary truck 60 m ahead. The driver braked and achieved a deceleration a = 5 m·s-2. What was the car’s braking distance? Did it hit the obstacle?

Solution:

Analysis:

v = 90 km·h-1 = 25 m·s-1, a = 5 m·s-2

 physics-kinematics-8

The braking distance of the car was s = 62.5 m. The car hit the obstacle.

 

9.A rocket reaches the second cosmic speed of 11 km·s-1 after traveling a distance of 200 km. How long does it take to achieve this? What is its acceleration?

Solution:

 

Analysis:

v = 11 km·s-1 = 11,000 m·s-1, s = 200 km = 200,000 m.

physics-kinematics-9

The rocket reaches the second cosmic speed in 36.4 s. Its acceleration is 302.5 m·s-2.

10.A cart on the rails of a demonstration set moves with constant acceleration a = 0.08 m·s-2 (v0 = 0, s0 = 0)

  • Calculate the distances the cart travels at times t = 1 s, 2 s, 3 s, 4 s, 5 s.
  • Determine the distances traveled in each consecutive second. In what ratio are these distances?

Solution:

physics-kinematics-10

11.The engineer of an express train, which was moving at v1 = 108 km·h-1, saw a freight train 180 m ahead moving in the same direction at v2 = 32.4 km·h-1. The engineer began braking and the train decelerated at a = 1.2 m·s-2. Determine whether the trains will collide.

Solution:

Analysis:

s0 = 180 m, v1 = 108 km·h-1 = 30 m·s-1, v2 = 32.4 km·h-1 = 9 m·s-1, a = 1.2 m·s-2

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 The trains will collide 15 s after braking begins, at a distance of 315 m.

12.The second hand of a clock is one third longer than the minute hand. In what ratio are the speeds of their tips?

Solution:

physics-kinematics-12

13.The Earth orbits the Sun in approximately uniform circular motion in 365.25 days. What is the speed of the Earth if the Earth–Sun distance is approximately 150 million kilometers?

Solution:

T = 365.25 days = 31,557,600 s = 3.16·107 s, r = 150,000,000 km = 1.5·1011 m.

physics-kinematics-13 

The Earth orbits the Sun with a tangential speed of 30 km·s-1.

14.The speed of uniform circular motion of a satellite around the Earth is 7.46 km·s-1. The satellite moves at an altitude of 800 km above the Earth’s surface. (R = 6378 km) Determine the orbital period T of the satellite around the Earth.

Solution:

R = 6378 km = 63.78·105 m, h = 800 km = 8·105 m, v = 7.46·103 m·s-1

r = R + h = 63.78·105 m + 8·105 m = 71.78·105 m


physics-kinematics-14

 

15.What is the minimum speed a motorcyclist must have to ride in all directions on the inner surface of a hollow sphere with radius R = 6 m? The center of mass of the motorcycle and rider is 0.9 m from the surface.

Solution:

r = R – 0.9 = 6 m – 0.9 m = 5.1 m

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The minimum speed of the motorcyclist must be 25.56 km·h-1.

 

16. A carousel seat is attached at a distance of 240 cm from the center of rotation and makes 18 revolutions per minute. Determine its tangential speed and centripetal acceleration.

Solution:

Analysis:

r = 240 cm = 2.4 m, f = 18/60 s-1 = 0.3 s-1,

a)   

v = 2π·r·f

v = 6.28·2.4 m·0.3 s-1 = 4.52 m·s-1

v = 4.52 m·s-1

 

b)

a c = 4π2·f2·r

a c = 4·9.8596·0.32·2.4 m·s-2 = 8.52 m·s-2

a c = 8.52 m·s-2

 

The carousel seat has a tangential speed of 4.52 m·s-1

and a centripetal acceleration of 8.52 m·s-2.