All categories

2 years ago

Average velocity is different than average speed because calculating average velocity involves

Physics

1 answer:

Artemon [7]2 years ago

8 0

Answer:

calculating displacement.

Explanation:

It's not true that displacement and distance would be the same always. Displacement is always smaller than or equal to distance as it is the smallest path between the initial and final point whereas distance is the measure of the total path covered.

You might be interested in

A car travels 40 miles in 30 minutes.

lukranit [14]

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 $m/s^2$

<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 = $\frac{1}{60}hours$

30 minute = $\frac{30}{60}hours$

30 minute = $\frac{1}{2}hours$

Now

Average velocity = $\frac{Speed}{time}$

Substituting Values,

Average velocity = $\frac{64.3738}{\frac[1}{2}}$

Average velocity = $64.3738 \times 2$

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

$128.74 \times \frac{5}{18}$

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

kinetic energy K= $\frac{1}{2}mv^2$

Substituting the values,

K= $\frac{1}{2}1500(35.75)^2$

K= $\frac{1}{2}1500(1278.06)$

K= $\frac{1500 \times (1278.06)}{2}$

K= $\frac{1917093.75}{2}$

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,

$S= ut+\frac{1}{2}at^2$

Substituting the known values,

$s= ut+\frac{1}{2}at^2$

$s= (37.75)(15)+\frac{1}{2}a(15)^2$

$s=566.25+\frac{1}{2}a(225)$

$s=566.25+\frac{(225a)}{2}$ -------------------------------------(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,

$a = \frac{v - u}{t}$

Substituting the values

$a = \frac{0 - 35.75}{15}$

$a = \frac{- 35.75}{15}$

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

Now substituting 4 in 3 we get

$s=566.25+\frac{(225( - 2.38)}{2}$

$s=566.25+\frac{-535.5}{2}$

$s=536.25-267.75$

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

4 0

3 years ago

Which section of the graph represents negative acceleration?

agasfer [191]

The area between the 10 and the 12.

8 0

3 years ago

Read 2 more answers

Which question cannot be answered through making measurements?

Bond [772]

A. is the answer for this question

5 0

3 years ago

Juan measured the temperature of salt water. He then added 273 to the measured value. Which conversion is Juan most likely doing

Semmy [17]

If Juan used a Celsius thermometer, it would tell him the Celsius temperature.

If he added 273 to that number, he'd have the "absolute" or Kelvin temperature.

7 0

3 years ago

A man wishes to lift a stone weighing 1440 N, using a first-class lever that measures 5 meters. What force should it perform if

Crazy boy [7]

Answer:

3360 N

Explanation:

In a first-class lever, the effort force and load force are on opposite sides of the fulcrum.

The lever is 5 m long. The load force is 1.50 m from the fulcrum, so the effort force must be 3.50 m from the fulcrum.

The torques are equal:

Fr = Fr

(1440 N) (3.5 m) = F (1.5 m)

F = 3360 N

4 0

3 years ago