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3 years ago

A body is accelerated continuously. What is the form of the graph?

1 answer:

6 0

That depends on what quantity is graphed.

It also depends on what kind of acceleration is taking place ...
continuous change of speed or continuous change of direction.

-- If the graph shows speed vs time, and the acceleration is a change
in speed, then the graph is a connected series of straight-line pieces.
Each straight piece slopes up if speed is increasing, or down if speed
is decreasing.

-- If the graph shows speed vs time, and the acceleration is a change in
direction only, then the graph is a straight horizontal line, since speed is
constant.

-- If the graph shows direction vs time, and the acceleration is a change
in speed only, then the graph is a straight horizontal line, since direction
is constant.

-- If the graph shows direction vs time, and the acceleration is a change
in direction, then the graph is a connected series of pieces of line.
Each piece may be straight if the direction is changing at a constant rate,
or curved if the direction is changing at a rate which grows or shrinks.
Each piece may slope up if the angle that defines the direction is growing,
or may slope down if the angle that defines the direction is decreasing.

-- If the graph shows distance vs time, and the acceleration is a
change in speed, then the graph is a connected series of pieces
of curves. Each piece curves up if speed is increasing, or down if
speed is decreasing.

-- If the graph shows distance vs time, and the acceleration is a change
in direction only, then the graph is a straight line sloping up, since speed
is constant.

You might be interested in

a particle is moving with shm of period 8.0s and amplitude 5.0cm. find (a) the speed of particle when it is 3.0m from the centre

Fudgin [204]

Answer:

a) $speed=\pi cm/s$

b) $v_{max}=\frac{5\pi}{4} cm/s$

c) $a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

Explanation:

The very first thing we must do in order to solve this problem is to find an equation for the simple harmonic motion of the given particle. Simple harmonic motion can be modeled with the following formula:

$y=Asin(\omega t)$

where:

A=amplitude

$\omega$ = angular frequency

t=time

we know the amplitude is:

A=5.0cm

and the angular frequency can be found by using the following formula:

$\omega=\frac{2\pi}{T}$

so our angular frequency is:

$\omega=\frac{2\pi}{8s}$

$\omega=\frac{\pi}{4}$

so now we can build our equation:

$y=5sin(\frac{\pi}{4} t)$

we need to find the speed of the particle when it is 3m from the centre of its motion, so we need to find the time t when this will happen. We can use the equation we just found to get this value:

$y=5sin(\frac{\pi}{4} t)$

$3=5sin(\frac{\pi}{4} t)$

so we solve for t:

$sin(\frac{\pi}{4} t)=\frac{3}{5}$

$\frac{\pi}{4} t=sin^{-1}(\frac{3}{5})$

$t=\frac{4}{\pi}sin^{-1}(\frac{3}{5})$

you can directly use this expression as the time or its decimal representation:

t=0.81933

since we need to find the speed of the particle at that time, we will need to get the derivative of the equation that represents the particle's position, so we get:

$y=5sin(\frac{\pi}{4} t)$

$y'=5cos(\frac{\pi}{4} t)*\frac{\pi}{4}$

which simplifies to:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

and we can now substitute the t-value we found previously, so we get:

$y'=\frac{5\pi}{4}cos(\frac{\pi}{4} (0.81933))$

$y'=\pi$

so its velocity at that point is $\pi$ cm/s

b) In order to find the maximum velocity we just need to take a look at the velocity equation we just found:

$y' =\frac{5\pi}{4}cos(\frac{\pi}{4} t)$

its amplitude will always give us the maximum velocity of the particle, so in this case the amplitude is:

$A=\frac{5\pi}{4}$

so:

$v_{max}=\frac{5\pi}{4} cm/s$

c) we can use a similar procedure to find the maximum acceleration of the particle, we just need to find the derivative of the velocity equation and determine its amplitude. So we get:

$y'= \frac{5\pi}{4}cos(\frac{\pi}{4} t)$

We can use the chain rule again to find this derivative so we get:

$y" =-\frac{5\pi}{4}sin(\frac{\pi}{4} t)*(\frac{pi}{4})$

so when simplified we get:

$y"=-\frac{5\pi^{2}}{16}sin(\frac{\pi}{4} t)$

its amplitude is:

$A=\frac{5\pi^{2}}{16}$

so its maximum acceleration is:

$a_{max}=\frac{5\pi^{2}}{16} cm/s^{2}$

7 0

3 years ago

If you are driving at a constant velocity of 5 m/s what is your<br> acceleration?

Paraphin [41]

"Acceleration" means the rate at which velocity is changing.

If velocity is constant, then it isn't changing.

If velocity isn't changing, then acceleration is <em>zero</em>.

3 0

3 years ago

A 0.58 kg mass is moving horizontally with a speed of 6.0 m/s when it strikes a vertical wall. The mass rebounds with a speed of

Sunny_sXe [5.5K]

Answer:

$5.8\; {\rm kg\cdot m \cdot s^{-1}}$ .

Explanation:

If the mass of an object is $m$ and the velocity of that object is $v$ , the linear momentum of that object would be $m\, v$ .

Assume that the initial velocity of the mass is positive ( $6.0\; {\rm m\cdot s^{-1}}$ .) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be $(-4.0\; {\rm m\cdot s^{-1}})$ .

Thus, the change in the velocity of the mass would be:

$\begin{aligned}& (\text{Change in Velocity}) \\ =\; & (\text{Final Velocity}) - (\text{Initial Velocity}) \\ =\; & (-4.0\; {\rm m\cdot s^{-1}}) - (6.0\; {\rm m\cdot s^{-1}}) \\ =\; & (-10\; {\rm m\cdot s^{-1})\end{aligned}$ .

The change in the linear momentum of the mass would be:

$\begin{aligned} & \text{change in momentum} \\ =\; & (\text{mass}) \times (\text{change in velocity}) \\ =\; & 0.58\; {\rm kg} \times (-10\; {\rm m\cdot s^{-1}}) \\ =\; & (-5.8\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}$ .

Thus, the magnitude of the change of the linear momentum would be $5.8\; {\rm kg \cdot m \cdot s^{-1}}$ .

7 0

3 years ago

In which example would the most transfer of energy take place from Mechanical energy to Thermal (heat) energy?

drek231 [11]

Answer:The answer is when a metal plate is pressed against a fast spinning piece of steel hard enough to stop it. i know this because I just took a test and got this question and got it write

Explanation:

5 0

3 years ago

How will the electrostatic force between two electric charges change if the first charge is doubled and the second charge is onl

Vinvika [58]

Answer:

B) $\frac{2}{3}$

Explanation:

The electric force between charges can be determined by;

F = $\frac{kq_{1} q_{2} }{r^{2} }$

Where: F is the force, k is the Coulomb's constant, $q_{1}$ is the value of the first charge, $q_{2}$ is the value of the second charge, r is the distance between the centers of the charges.