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Mama L [17]
2 years ago
7

How do you find distance from average velocity and time

Physics
1 answer:
AlekseyPX2 years ago
7 0

Answer:

Calculate the total distance travelled by the object - its motion is represented by the velocity-time graph below.

Here, the distance travelled can be found by calculating the total area of the shaded sections below the line.

½ × base × height.

½ × 4 × 8 = 16 m 2

(10 – 4) × 8 = 48 m 2

Explanation:

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An object is released from rest and falls a distance h during the first second of time. How far will it fall during the next sec
Viefleur [7K]

Answer:

E. 3h

Explanation:

We know that

u = 0 m/s.

velocity after t = 1s

v = u+gt = 0+9.81 x 1s= 9.81 m/s

distance covered in 1st sec

= =>> ut+0.5 x g x t²

=>>0 + 0.5x 9.81 x 1 = 4.90m

Let 4.90 be h

distance travelled in 2nd second will now be used

So velocity after t = 1s

=>>1 x t+ 0.5 x g x t²

=>9.81x 1 + 0.5 x 9.81 x 1 = 3 x 4.90

So since h= 4.90

Then the ans is 3x h = 3h

3 0
2 years ago
What was the direction of the ball’s velocity
Tju [1.3M]

Part of the question is missing. Here it is:

<em>A 72 g autographed baseball slides off of a 1.3 m high table and strikes the floor a horizontal distance of 0.7m away from the table.     The acceleration of gravity is 9.81 m/s2. What was the direction of the ball’s velocity  just before it hit the floor? </em>

Answer:

\theta=-75.7^{\circ}

Explanation:

The motion of the ball is a projectile motion, which consists of two separate motions:

- A horizontal motion at constant velocity

- A vertical motion at constant acceleration (free fall)

We start by analyzing the vertical motion, to find the time of flight of the ball. This can be done by using the suvat equation

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

where, choosing downward as positive direction:

s =1.3 m is the vertical displacement of the ball

u = 0 is the initial vertical velocity

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

t is the time

Solving for t,

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(1.3)}{9.8}}=0.52 s

Now we can find the final vertical velocity of the ball, using:

v_y=u+at

And susbtituting t = 0.52 s, we find

v_y = 0 +(9.8)(0.52)=5.1 m/s

It is important to keep in mind that the direction of this velocity is downward, since we chose downward as positive direction.

The horizontal velocity of the ball instead is constant; we know that the ball covers a horizontal distance of

d = 0.7 m

In a time of

t = 0.52 s

So, the horizontal velocity is

v_x = \frac{0.7}{0.52}=1.3 m/s

So now we can find the direction of the ball's velocity using:

\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{5.1}{1.3})=75.7^{\circ}

And since the vertical direction is downward, this means that this velocity is below the horizontal, so the answer is

\theta=-75.7^{\circ}

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Suppose you are on a cart that is moving at a constant speed v toward the left on a frictionless track. If you throw a massive b
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