The speed of an object undergoing constant acceleration increases from 8.0 m/s to 12.0 m/s in 10.0 seconds.

6 answers
Question:

The speed of an object undergoing constant acceleration increases from 8.0 m/s to 12.0 m/s in 10.0 seconds. How far does the object travel during the 10.0 seconds?
4 m
100 m
800 m
10 m

Answers

is the question finnished

Explanation:

687.5 m

Explanation:

as the given speed (or velocity) is given as 12.5 m/s and time is 55 sec...

then total distance to be travelled in 55 secs= (12.5× 55) m = 687.5 m

:)

It traveled 4 centimeters.

Explanation:

In a speed versus time graph, the distance travelled is given by the area under the graph.

In this graph we have the following:

- The speed of the object is v = 1 cm/s between time t = 0 s and t = 4 s

- The speed of the object is v = 0 cm/s between time t = 4 s and t = 8 s

Since the speed in the second part is zero, the distance travelled in the second part is zero. So, the only distance travelled by the object is the distance travelled during the first part, which is equal to the area of the first rectangle:

[tex]d=v\Delta t=(1)(4-0)=4 cm[/tex]

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The answer to your question is 24 cm

Explanation:

The travel is divided into two sections:

1.- From 0 to 4 seconds

2.- From 4 to 8 seconds

In the first travel speed changes and on the second travel, speed is constant.

1.- Distance 1

   [tex]a = \frac{4 - 0}{4}[/tex]

          a = 1 cm/s²

   [tex]d = \frac{v^{2}f - v^{2}o }{2a}[/tex]

   [tex]d = \frac{4^{2} - 0^{2} }{2(1)}[/tex]

   [tex]d = \frac{16}{2}[/tex]

          d = 8 cm

2.- Distance

    [tex]speed = \frac{distance}{time}[/tex]

           distance = speed x time

           distance = (4) x 4

           distance = 16 cm

3.- Total distance

           d = 8 + 16

          d = 24 cm

         

yes

Explanation:

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