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Chrissyo
What would the x vs t graph look like with the following Negative displacement, positive velocity, and negative acceleration
Hi! Are these different graphs you're looking for? Or should one graph describe the all?
I assume one graph to describe them all, @Chrissyo ?
i just need the xt graph
the vt is a constant downward line in the 4th quad. and the at is negative constant accel straight line
First let's just consider the meaning of the signs (negative and positive, I mean). Whe you create this theoretical physics world of yours every time you start a problem, you're working in two dimensions - for now. AND one direction is positive and one direction is negative. Now let's think about what t, or time, is in relation to our physics problem/world. We just say that the start of the event we're looking at happens at t=0. Like in a free fall problem where we let go of a ball, we let go at t = 0. Why not? It's very convenient. Let's recall the meanings of displacement, velocity, and acceleration, now: Displacement is distance from one specified point to another. Now it's from the starting location point to the location of the object at the time, t, we look at. It describes whether you're more foward (positive) or behind (negative) the initial point. Velocity is a change in displacement as time goes on. It's necessary to move forwards (positive) or move backwards (negative). Acceleration is a change in the velocity over time. It's necessary to speed up forwards(positive) or speed up backwards (negative). If you velocity is forwards, then a backwards acceleration will slow velocity to a point of stopping, and, afterwards, velocity will be backwards, too.
Negative displacement: behind the initial starting point Positive velocity: traveling fowards (here, towards the starting point) Negative acceleration: speeding up backwards (here, slowing to a stop, and then heading backwards)
Start with the most complex first. Acceleration|dw:1350777626072:dw| Velocity, changing because of acceleration|dw:1350777653573:dw| Displacement, changing at a rate set by velocity|dw:1350777741982:dw| and displacement might start going down, which still satisfies your problem.|dw:1350777996893:dw| If this problem actually describes 3 situations, I hope my little tutorial has helped you! :) Feel free to message me or respond and even tag me! Good luck!