Answer:
(a)Average velocity ,v =128.74 Km/hr
(b)Kinetic Energy , K=958546.875 Joule
(c)Distance, s=268.8m
(d)Acceleration, a= - 2.38 ![m/s^2](https://tex.z-dn.net/?f=m%2Fs%5E2)
<u>Explanation</u>:
<u>Given</u>:
Distance travelled = 40 miles
Time taken = 30 minutes.
(A) The average velocity in kilometres/hour
Converting 40 miles into km ,
we know that,
1 mile = 1.60934
40 miles = 40 x 1.60934
so 40 miles = 64.3738 Km
similarly converting 30 minutes into hours
1 minute = ![\frac{1}{60}hours](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B60%7Dhours)
30 minute = ![\frac{30}{60}hours](https://tex.z-dn.net/?f=%5Cfrac%7B30%7D%7B60%7Dhours)
30 minute = ![\frac{1}{2}hours](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dhours)
Now
Average velocity = ![\frac{Speed}{time}](https://tex.z-dn.net/?f=%5Cfrac%7BSpeed%7D%7Btime%7D)
Substituting Values,
Average velocity = ![\frac{64.3738}{\frac[1}{2}}](https://tex.z-dn.net/?f=%5Cfrac%7B64.3738%7D%7B%5Cfrac%5B1%7D%7B2%7D%7D)
Average velocity = ![64.3738 \times 2](https://tex.z-dn.net/?f=64.3738%20%5Ctimes%202)
Average velocity =128.74 Km/hr
(B) If the car weighs 1.5 tons, what is its If the car weighs 1.5 tons, what is its kinetic energy in joules (Note: you will need to convert your velocity to m/s)? in joules (Note: you will need to convert your velocity to m/s)?
Converting 1.5 tons into kg we get
1 ton = 1000 kg
so 1.5 ton =1500 kg
converting velocity to m/s
![128.74 \times \frac{5}{18}](https://tex.z-dn.net/?f=128.74%20%20%5Ctimes%20%5Cfrac%7B5%7D%7B18%7D)
=>35.75 m/s----------------------------------------------------------(1)
kinetic energy K= ![\frac{1}{2}mv^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
Substituting the values,
K= ![\frac{1}{2}1500(35.75)^2](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D1500%2835.75%29%5E2)
K= ![\frac{1}{2}1500(1278.06)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D1500%281278.06%29)
K= ![\frac{1500 \times (1278.06)}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1500%20%5Ctimes%20%281278.06%29%7D%7B2%7D)
K= ![\frac{1917093.75}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1917093.75%7D%7B2%7D)
K=958546.875 Joule---------------------------------------------(2)
(c)When the driver applies the brake, it takes 15 seconds to stop. How far does the car travel (in meters) while stopping
Lets use Distance formula,
![S= ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=S%3D%20ut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
Substituting the known values,
![s= ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=s%3D%20ut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
![s= (37.75)(15)+\frac{1}{2}a(15)^2](https://tex.z-dn.net/?f=s%3D%20%2837.75%29%2815%29%2B%5Cfrac%7B1%7D%7B2%7Da%2815%29%5E2)
![s=566.25+\frac{1}{2}a(225)](https://tex.z-dn.net/?f=s%3D566.25%2B%5Cfrac%7B1%7D%7B2%7Da%28225%29)
-------------------------------------(3)
(D) What is the average acceleration of the car (in m/s2) during braking?
Using the formula
v=u +at
re arranging the formula we get,
![a = \frac{v - u}{t}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7Bv%20-%20u%7D%7Bt%7D)
Substituting the values
![a = \frac{0 - 35.75}{15}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B0%20-%2035.75%7D%7B15%7D)
![a = \frac{- 35.75}{15}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B-%2035.75%7D%7B15%7D)
a= - 2.38
----------------------------------------(4)
Now substituting 4 in 3 we get
![s=566.25+\frac{(225( - 2.38)}{2}](https://tex.z-dn.net/?f=s%3D566.25%2B%5Cfrac%7B%28225%28%20-%202.38%29%7D%7B2%7D)
![s=566.25+\frac{-535.5}{2}](https://tex.z-dn.net/?f=s%3D566.25%2B%5Cfrac%7B-535.5%7D%7B2%7D)
![s=536.25-267.75](https://tex.z-dn.net/?f=s%3D536.25-267.75)
s=268.8m--------------------------------------------------------------(5)