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 [tex]m/s^2[/tex]
Explanation:
Given:
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 = [tex]\frac{1}{60}hours[/tex]
30 minute = [tex]\frac{30}{60}hours[/tex]
30 minute = [tex]\frac{1}{2}hours[/tex]
Now
Average velocity = [tex]\frac{Speed}{time}[/tex]
Substituting Values,
Average velocity = [tex]\frac{64.3738}{\frac[1}{2}}[/tex]
Average velocity = [tex]64.3738 \times 2[/tex]
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
[tex]128.74 \times \frac{5}{18}[/tex]
=>35.75 m/s----------------------------------------------------------(1)
kinetic energy K= [tex]\frac{1}{2}mv^2[/tex]
Substituting the values,
K= [tex]\frac{1}{2}1500(35.75)^2[/tex]
K= [tex]\frac{1}{2}1500(1278.06)[/tex]
K= [tex]\frac{1500 \times (1278.06)}{2}[/tex]
K= [tex]\frac{1917093.75}{2}[/tex]
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,
[tex]S= ut+\frac{1}{2}at^2[/tex]
Substituting the known values,
[tex]s= ut+\frac{1}{2}at^2[/tex]
[tex]s= (37.75)(15)+\frac{1}{2}a(15)^2[/tex]
[tex]s=566.25+\frac{1}{2}a(225)[/tex]
[tex]s=566.25+\frac{(225a)}{2}[/tex]-------------------------------------(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,
[tex]a = \frac{v - u}{t}[/tex]
Substituting the values
[tex]a = \frac{0 - 35.75}{15}[/tex]
[tex]a = \frac{- 35.75}{15}[/tex]
a= - 2.38 [tex]m/s^2[/tex]----------------------------------------(4)
Now substituting 4 in 3 we get
[tex]s=566.25+\frac{(225( - 2.38)}{2}[/tex]
[tex]s=566.25+\frac{-535.5}{2}[/tex]
[tex]s=536.25-267.75[/tex]
s=268.8m--------------------------------------------------------------(5)
