Question

A car travels 40 miles in 30 minutes.

a. What is the average velocity in kilometers/hour?
b. 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)?
c. When the driver applies the brake, it takes 15 seconds to stop. How far does the car travel (in meters) while stopping?
d. What is the average acceleration of the car (in m/s2) during braking?

Answer 1

(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`

Step-by-step 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 =
`(1)/(60)hours`

30 minute =
`(30)/(60)hours`

30 minute =
`(1)/(2)hours`

Now

Average velocity =
`(Speed)/(time)`

Substituting Values,

Average velocity =
`(64.3738)/(\frac[1){2}}`

Average velocity =
`64.3738 * 2`

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  * (5)/(18)`

=>35.75 m/s----------------------------------------------------------(1)

kinetic energy K=
`(1)/(2)mv^2`

Substituting the values,

K=
`(1)/(2)1500(35.75)^2`

K=
`(1)/(2)1500(1278.06)`

K=
`(1500 * (1278.06))/(2)`

K=
`(1917093.75)/(2)`

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+(1)/(2)at^2`

Substituting the known values,

`s= ut+(1)/(2)at^2`

`s= (37.75)(15)+(1)/(2)a(15)^2`

`s=566.25+(1)/(2)a(225)`

`s=566.25+((225a))/(2)`-------------------------------------(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 = (v - u)/(t)`

Substituting the values

`a = (0 - 35.75)/(15)`

`a = (- 35.75)/(15)`

a= - 2.38
`m/s^2`----------------------------------------(4)

Now substituting 4 in 3 we get

`s=566.25+((225( - 2.38))/(2)`

`s=566.25+(-535.5)/(2)`

`s=536.25-267.75`

s=268.8m--------------------------------------------------------------(5)