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inna [77]
3 years ago
13

The position-time graph for a bug crawling along a line is shown in item 4 below. Determine whether the velocity is positive, ne

gative, or zero at each of the times marked on the graph.

Physics
1 answer:
Naddika [18.5K]3 years ago
5 0

Answer: The velocity at different marked time points are given as

t1 = -

t2 = +

t3 = +

t4 = -

t5 = 0

Explanation:

The slope of the tangent of the curve indicates the instantaneous velocity. So if the slope of the tangent is positive, that Is, the tangent makes a positive angle (above the horizontal axis) with the horizontal

axis, then the velocity at this point is positive, and if the slope of the tangent is negative, that is the tangent makes a negative angle with the horizontal axis (below the horizontal axis), then the velocity at this point is negative.

When the tangent of the line is parallel to the horizontal axis, the velocity is 0.

From the position-time graph attached, the sign on the instantaneous velocity for each time marked on the graph is given below

t1 = -

t2 = +

t3 = +

t4 = -

t5 = 0

QED!

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An Olympic track runner starts from rest and has an acceleration of 2.4 m/s2 for 3.6 s, then has zero acceleration for the remai
rjkz [21]

Answer:

The runner's speed at the following times would remain 8.64 m/s.

Explanation:

Acceleration definition: Acceleration is rate of change in velocity of an object with respect to time.

In this case, after 3.6 seconds the acceleration is zero, it means that the velocity of the runner after 3.6 seconds is not changing and it will remain constant for the remainder of the race. Now, we have to find the velocity of the runner that he had after 3.6 seconds and that would be the runner's speed for the remainder of the race. For this we use first equation of motion.

First equation of motion:        Vf = Vi + a×t

Vf stands for final velocity

Vi stands for initial velocity

a stands for acceleration

t stands for time

In the question, it is mentioned that the runner starts from rest so its initial velocity (Vi) will be 0 m/s.

The acceleration (a) is given as 2.4 m/s²

The time (t) is given as 3.6 s

Now put the values of Vi, a and t in first equation of motion

                       Vf = Vi + a×t

                       Vf = 0 + 2.4×3.6

                       Vf = 2.4×3.6

                       Vf = 8.64 m/s

So,the runner's speed at the following times would remain 8.64 m/s.

5 0
3 years ago
PLEASE HELP!!! GIVING BRAINLIEST!! ill also answer questions that you have posted if you answer these correctly!!!! (40pts)
Cerrena [4.2K]

Answer:

B false

Explanation:

dont have any explanation

8 0
3 years ago
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