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3 years ago

If a distance-time graph contains the point (4, 15), what does that tell you about the runner?

Physics

1 answer:

serg [7]3 years ago

5 0

Explanation:

It is given that there is distance time graph. The slope of any distance time graph will give the speed or velocity of the object.

In the context, it is seen that there is a point in the distance - time graph. And the values of the point in the graph is (4,15).

Now here, the distance is plotted in the ordinate or the vertical y-axis while time is plotted in the abscissa or the horizontal x-axis.

Thus the point (4,15) tells us that the object has traveled 15 units in time equal to 4 seconds.

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3 years ago

A student throws a rock horizontally from the edge of a cliff that is 20 m high. The rock has an initial speed on 10 m/s. If air

fiasKO [112]

The distance of the rock from the base of the cliff is C) 20 m

Explanation:

The motion of the rock in this problem is a projectile motion, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction

We start by analyzing the vertical motion to find the time of flight of the rock (the time it takes to reach the ground). We can do it by using the suvat equation:

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

where, taking downward as positive direction,

s = 20 m is the vertical displacement of the rock

$u_y=0$ is the initial vertical velocity

t is the time of flight

$a=g=9.8 m/s^2$ is the acceleration of gravity

Solving for t,

$t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(20)}{9.8}}=2.02 s$

Now we can analzye the horizontal motion: the rock moves horizontally with a constant velocity of

$v_x = 10 m/s$

Therefore, the horizontal distance covered after a time t is

$d=v_x t$

and substituting t = 2.02 s, we find the final distance of the rock from the base of the cliff:

$d=(10)(2.02)=20 m$

Learn more about projectile motion:

brainly.com/question/8751410

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