Graphing Motion

You kick a ball with a speed of 14 m/s at an angle of 51°. How far away does the ball land?

In-s [12.5K]

-- The vertical component of the ball's velocity is 14 sin(51°) = 10.88 m/s

-- The acceleration of gravity is 9.8 m/s².

-- The ball rises for 10.88/9.8 seconds, then stops rising, and drops for the
same amount of time before it hits the ground.

-- Altogether, the ball is in the air for (2 x 10.88)/(9.8) = 2.22 seconds
==================================

-- The horizontal component of the ball's velocity is 14 cos(51°) = 8.81 m/s

-- At this speed, it covers a horizontal distance of (8.81) x (2.22) = 19.56 meters
before it hits the ground.

As usual when we're discussing this stuff, we completely ignore air resistance.

4 0

3 years ago

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A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co

algol [13]

Answer:

$\displaystyle X_T=66.6\ km$

Explanation:

Accelerated Motion

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

$\displaystyle V_f=V_o+a\ t$

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of $0.987 m/s^2$ for 182 seconds. Then it brakes with a constant acceleration of $-0.321 m/s^2$ until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

$\displaystyle X=V_o\ t+\frac{a\ t^2}{2}$

Our data is

$\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec$

Let's compute the first distance X1

$\displaystyle X_1=0+\frac{0.987\times 182^2}{2}$

$\displaystyle X_1=16,346.7\ m$

Now, we find the speed at the end of the first period of time

$\displaystyle V_{f1}=0+0.987\times 182$

$\displaystyle V_{f1}=179.6\ m/s$

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

$\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0$

$\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}$

$\displaystyle t=559.5\ sec$

Computing the second distance

$\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}$

$\displaystyle X_2=50,243.2\ m$

The total distance is

$\displaystyle X_t=x_1+x_2=16,346.7+50,243.2$

$\displaystyle X_t=66,589.9\ m$

$\displaystyle \boxed{X_T=66.6\ km}$

8 0

4 years ago

Visible light travels at a speed 3.0 × 108 of m/s. If red light has a wavelength of 6.5 × 10–7 m, what the frequency of this lig

gayaneshka [121]

i think its d on edge

4 0

4 years ago

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. Maria walked 1.5 miles south to her house in 0.5 hours. What is her speed in miles per hour?

Simora [160]

1) 3 miles/Hour

The speed is defined as the distance covered divided by the time taken:

$v=\frac{d}{t}$

where

d = 1.5 mi is the distance

t = 0.5 h is the time taken

Substituting,

$v=\frac{1.5}{0.5}=3 mi/h$

2) 1.34 m/s south

Velocity, instead, is a vector, so it has both a magnitude and a direction. We have:

$d=1.5 mi \cdot 1609 m/mi = 2414 m$ is the displacement in meters

$t=0.5 h \cdot 3600 s/h =1800 s$ is the time taken in seconds

Substituting,

$v=\frac{2414 m}{1800 s}=1.34 m/s$

And the direction of the velocity is the same as the displacement, so it is south.

6 0

4 years ago

The motion of an electron is given by x(t)=pt3+qt2+r, with p = -2.3 m/s^3 ,q = +1.5 m/s^2 , and r = +9.0 m.A) Determine its velo

alexandr402 [8]

Answer:

$v(0)=0\\v(1)=-3.9\ m/s\\v(2)=-21.6\ m/s\\v(3)=-53.1\ m/s$

Explanation:

Instant Velocity

Given the position as a function of time x(t), the instant velocity is the derivative of the function:

$v(t)=x'(t)$

We are given the position as

$x(t)=-2.3t^3+1.5t^2+9$

The derivative of x is

$v(t)=x'(t)=-6.9t^2+3.0t$

A) Let's compute v(0)

$v(0)=-6.9(0)^2+3.0(0)=0$

B)

$v(1)=-6.9(1)^2+3.0(1)$

$v(1)=-3.9\ m/s$

C)

$v(2)=-6.9(2)^2+3.0(2)$

$v(2)=-21.6\ m/s$

D)

$v(3)=-6.9(3)^2+3.0(3)$

$v(3)=-53.1\ m/s$

4 0

3 years ago