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maksim [4K]
4 years ago
13

I WILL MARK BRAINIEST!!

Physics
2 answers:
djyliett [7]4 years ago
6 0

Answer:

as the slope of the line

Explanation:

As we know that the speed is defined as the rate of distance moved by the object.

Here let say that total distance that is moved by the object is "d" and total time taken by the object is "t"

so here average speed is given as

v = \frac{d}{t}

now we say that for small interval of time or for an instant of time the speed is given as

v = \frac{dx}{dt}

so here it shows that it is slope of distance time graph so speed is the slope of the graph at given point

Ivahew [28]4 years ago
3 0
The answer is as the slope of the line
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A projectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal compon
Anettt [7]
First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed v_x=30 m/s, and an accelerated motion on the y-axis, with initial speed v_y=20 m/s and acceleration g=9.81 m/s^2:
S_x(t)=v_xt
S_y(t)=v_y t- \frac{1}{2} gt^2
where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).

To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring
S_y(t)=0
Therefore:
v_y t -  \frac{1}{2}gt^2=0
which has two solutions:
t=0 is the time of the beginning of the motion,
t= \frac{2 v_y}{g} = \frac{2\cdot 20 m/s}{9.81 m/s^2}=4.08 s is the time at which the projectile hits the ground.

Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
S_x(4.08 s)=v_x t=(30 m/s)(4.08 s)=122.4 m
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aksik [14]

Answer:

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hope this helps

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