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Zielflug [23.3K]
3 years ago
15

1. How would you expect an instantaneous acceleration vs. time graph to look for a cart moving with a constant

Physics
1 answer:
r-ruslan [8.4K]3 years ago
6 0

Answer:

a=0   v = v₀ + a t

a=0    line is horizontal

Explanation:

1, In a graph of acceleration vs. time, we have lines, when the line is horizontal it is zero, when the line has a positive slope the increasing accelerations and when the slope is negative the decreasing acceleration

2, speed and relationship of a car is given by

        v = v₀ + a t

where vo is the initial velocity, a is the acceleration and tel time

in this case I will calcograph velocity vs. time the constant acceleration is a straight line.

In general from the graph we can find the initial velocity with the cut at that x and the acceleration of the car with the slope

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Answer:

-5.8868501529 m/s² or -5.8868501529g

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Explanation:

t = Time taken

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Dividing by g

\dfrac{a}{g}=\dfrac{-57.75}{9.81}\\\Rightarrow a=-5.8868501529g

The acceleration is -5.8868501529 m/s² or -5.8868501529g

v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{1.3-2}{-5.8868501529}\\\Rightarrow t=0.118909090909\ s

The time taken is 0.118909090909 s

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