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Figure shows several graphs with unlabelled axes.(a) Which graph would best represent velocity as a function of time for an object moving with constant speed?(b) Which graph best represents velocity as a function of time for acceleration given by a = +3t? (c) Which graph best represents distance as a function of time for a constant negative acceleration?(d) Which graph best represents velocity as a function of time if graph E shows distance as a function of time?

Amit Saxena , 10 Years ago
Grade upto college level
anser 1 Answers
Deepak Patra

Last Activity: 10 Years ago

(a)

The correct option is: B
(b)
The correct option is: D
(c)
The correct option is: E
(d)
The correct option is: C
(a)
The object moving with a constant speed can change the direction of the velocity vector but not its magnitude, and therefore the graph that best represents the velocity as a function of time is:
233-213_Untitled.png
It can be seen from the figure that the magnitude of the velocity remains constant at all times.
(b)
The velocity time curve can be obtained from the acceleration as:
V = \int adt
Substitute , a = +3t
v = \int 3tdt
= 3\int tdt |
= \frac{3t^{2}}{2} + c
Here, C is a constant of integration.
It can be seen from equationv = \frac{3t^{2}}{2} + C that velocity depends on the second power of variable t and should therefore generate a parabola. However, the positive acceleration ensures that the change in velocity in itself remains positive, and the final velocity is greater than the initial velocity in magnitude.
Therefore, the velocity time graph corresponding to acceleration a = +3t is:
233-1540_1.png
It can be seen from the figure above that the slope of the function increases linearly with time.
(c)
Assume that the acceleration of the object is a = -k ( k being a constant value). The velocity corresponding to acceleration a = -k is:
v = \int -kdt
= - k \int dt
= - kt + C
Here, C is a constant of integration.
The position corresponding to acceleration a = – k is:
x = \int vdt
Substitute ,v = kt + C,
x = \int (-kt+C) dt
= \frac{-kt^{2}}{2} + Ct + C_{1}
Here, is C1 a constant of integration.
It can be seen from equationx = \frac{-kt^{2}}{2}+ Ct + C_{1} that the position depends both on the first and second power of variable t , and therefore the graph representing the same should be a parabola.
However, graph represented by D corresponds to a positive slope for positive value t of whereas graph E represented to negative slope for positive values of t and is in accordance with function . x = \frac{-kt^{2}}{2}+ Ct + C_{1}.
Therefore, the position time graph corresponding to constant negative acceleration is:
233-1321_1.png
It can be seen from the figure above that at t = 0 and t \rightarrow \infty, the magnitude of velocity increases infinitely.
(d)
The slope of the distance time graph represented by E at a point P is zero. Therefore, the velocity of the object at same point P should be a constant value and vary symmetrically about that point . Therefore, the velocity time graph will have positive values before point P and negative values after point P. This condition is fulfilled only by graph C .

233-906_1.png
It can be seen in the figure below that the line represented by C has positive values at first, zero at point P and negative thereafter.

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