Physics, asked by rishitpandey77, 1 year ago

show the graph of constant acceleration on v-t graph​

Answers

Answered by Anonymous
1

Answer:

Position-time (x-t) graph

x=x0+v0t+12at2

The graph position-time (x-t) of a constant acceleration motion, or uniformly accelerated rectilinear motion (u.a.r.m.), represents time on the horizontal axis (t-axis) and position on the vertical axis (x-axis). Observe as the position (normally the x-coordinate) increases (or decreases) uniformly with time. This happens because as time passes, the magnitude of the velocity varies. We can distinguish two cases, when the velocity is positive or when it is negative:

Position - time (x-t) graph in constant acceleration motion

Velocity-time (v-t) graph

v=v0+a⋅t

The graph velocity-time (v-t) of a constant acceleration motion, or uniformly accelerated rectilinear motion (u.a.r.m.), represents time on the horizontal axis (t-axis) and velocity on the vertical axis (x-axis). Observe as the velocity increases (or decreases) uniformly with the passage of time. This happens as the result of the acceleration. Again, we can distinguish two cases:

Velocity - time (v-t) graph in constant acceleration motion

We can get the velocity from the angle α. To do it just remember that in a right triangle the tangent of each of its angles is defined as the opposite side (cathetus) divided by the adjacent one:

tanα=opposite sideadjacent side=ΔvΔt=v−v0t=a

The value of the slope is the magnitude of the acceleration. Therefore, the greater the slope of the straight line, the higher the acceleration of the body.

Notice that the area under the curve v between two instants of time, numerically matches the distance traveled. Could you tell why?

Distance traveled and area under the velocity graph in constant acceleration motion

The area under the curve can be calculated as the area of the rectangleo S1 that would correspond to a uniform rectilinear motion (u.r.m.) to which we will add the area of the triangle S2:

Δx=x−x0=S1+S2=1v0t+(v−v0)t2=2v0t+12at2

Where have we applied:

{S1=v0tS2=Srectangle2=(v−v0)t2

v−v0=at

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