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Ivanshal [37]
4 years ago
14

Sally Leadfoot was pulled over on her way from Syracuse to Ithaca by an officer claiming she was speeding. The speed limit is 65

mi/hr and Sally had traveled 97 km in 102 minutes. How fast was Sally’s average speed? Does she deserve a ticket?
Physics
1 answer:
Citrus2011 [14]4 years ago
3 0

Sally's average speed is <u>35.3 mi/h.</u>

Average speed of a body is the total distance traveled in the given time interval.

Express the distance  d traveled in miles.

d=\frac{97 km}{1.6 km/m\\i} \\ = 60 mi

Express the time t traveled in hours.

t  =\frac{102 min}{60 min/hr}\\ = 1.7 h

Calculate the average speed v.

v =\frac{d}{t} \\ =\frac{60 mi}{1.7h} \\   = 35.3 mi/h

Her average speed is 35.3 mi/h, which is less than the speed limit of 65 mi/h.

However, the average speed of an object is different from its instantaneous speed. It could be possible that at the time when the officer apprehended her, Sally could have been travelling at a speed greater than the prescribed speed limit, which would have prompted the officer to issue a speeding ticket to her.

Thus, the average speed of a person cannot be considered as a bench mark for speeding offences, since her instantaneous speed could have been higher than the speeding limit and yet she could have had an average speed less than the speeding limit.


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an object was accelerated with a force of 150 N from rest to a velocity of 10m/s in 5 sec. the force of friction opposing the ob
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7 0
3 years ago
A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho
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Answer:

Part a)

KE_r = 8 J

Part b)

v = 3.64 m/s

Part c)

KE_f = 12.7 J

Part d)

v = 2.9 m/s

Explanation:

As we know that moment of inertia of hollow sphere is given as

I = \frac{2}{3}mR^2

here we know that

I = 0.0484 kg m^2

R = 0.200 m

now we have

0.0484 = \frac{2}{3}m(0.200)^2

m = 1.815 kg

now we know that total Kinetic energy is given as

KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2

20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2

20 = 1.5125 v^2

v = 3.64 m/s

Part a)

Now initial rotational kinetic energy is given as

KE_r = \frac{1}{2}I(\frac{v}{R})^2

KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2

KE_r = 8 J

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

mgh = (KE_i) - KE_f

1.815(9.8)(0.900sin27.1) = 20- KE_f

7.30 = 20 - KE_f

KE_f = 12.7 J

Part d)

Now we know that

\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7

\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7

1.5125 v^2 = 12.7

v = 2.9 m/s

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