Which graph illustrates constant speed and velocity?

Physics

2 answers:

boyakko [2]3 years ago

8 0

The correct graph is <u>D</u>.

The graph <em>A</em> is a straight line sloping downwards and it shows that the speed of the body is decreasing at a constant rate. Therefore, this s a graph of a body that is under a constant deceleration.

The graph B is a straight line which slopes upwards. Hence the graph shows that the speed of the body increases at a constant rate. Therefore, this is a graph of a body that is accelerating at a constant rate.

The graph C is curved line, which curves upwards. The slope of the curve increases with time. This is therefore, a graph of a body which is under increasing acceleration.

The graph D, however is a straight line parallel to the time axis. The speed of the body has the same value at all times. Therefore, Graph D is the graph which shows the motion of a body with constant speed.

denpristay [2]3 years ago

5 0

Answer choice D is correct.

When you are thinking about a "constant" on a graph, it usually means that is is a straight line that is either horizontal or vertical. In this case the only graph that as both a constant speed and velocity is D which is why it is correct.

A has a negative constant acceleration

B has a positive acceleration

C is not constant, but is moving at a positive acceleration.

Hope this helps some!

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4 years ago

The total energy of a block—spring system is 0.18 J. The amplitude is 14.0 cm and the maximum speed is 1.25 m/s. Find: (a) the m

algol13

a) The mass is 0.23 kg

b) The spring constant is 1.25 N/m

c) The frequency is 1.42 Hz

d) The speed of the block is 1.08 m/s

Explanation:

We can find the mass of the block by applying the law of conservation of energy: in fact, the total mechanical energy of the system (which is sum of elastic potential energy, PE, and kinetic energy, KE) is constant:

$E=PE+KE=const.$

The potential energy is given by

$PE=\frac{1}{2}kx^2$

where k is the spring constant and x is the displacement. When the block is crossing the position of equilibrium, x = 0, so all the energy is kinetic energy:

$E=KE_{max}=\frac{1}{2}mv_{max}^2$ (1)