Acceleration of Velocity Graph: Finding the Relationship | Worsley School

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AI Thread Summary
The discussion centers on interpreting a velocity graph to determine acceleration. It is established that the acceleration is zero for the entire duration of the graph, as the velocity remains constant except for instantaneous changes at specific points. If there were non-zero acceleration, the graph would exhibit a slope, indicating changes in velocity over time. The conclusion confirms that the acceleration graph would indeed be a straight line along the t-axis, representing zero acceleration. This understanding clarifies the relationship between velocity and acceleration in the given context.
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Homework Statement




http://www.worsleyschool.net/science/files/average/graph1b.gif


Is the acceleration zero for this entire velocity graph?


Homework Equations



v = delta x/ delta t
a= delta v/ delta t

The Attempt at a Solution



If i were to convert this velocity graph to an acceleration graph, i think it would be a straight line on the t-axis.
 
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yes, the velocity is constant all the time, it just changes instantaneously at t=3 and at t=6. If there were non-zero acceleration the velocity would change over time and you would see a slope (upwards or downwards) in a velocity versus time graph.
 
gerben said:
yes, the velocity is constant all the time, it just changes instantaneously at t=3 and at t=6. If there were non-zero acceleration the velocity would change over time and you would see a slope (upwards or downwards) in a velocity versus time graph.

thank you very much for the reply,

but just to make sure: so the acceleration graph of this velocity graph would be just a straight line on the t-axis ( 0 acceleration) ?
 
yes that's correct
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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