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(a) Sketch a graph of velocity versus time corresponding to the graph of displacement versus time given in Figure 2.55.

(b) Identify the time or times ( ta , tb , tc , etc.) at which the instantaneous velocity is greatest.

(c) At which times is it zero?

(d) At which times is it negative?

Short Answer

Expert verified

a.

b) The time at which instantaneous velocity is greatest is td.

c) Instantaneous velocity is zero at tc, te, tg, tl .

d) The velocity is negative at ta, tb, and tf.

Step by step solution

01

Velocity vs. Time Graph  

The velocity vs. time graph we get when we take the slope of the displacement versus time graph.

The slop of this graph gives the value of the velocity, and it also determines the direction of the velocity

V=³Ù²¹²Ôθ

If we try to draw the slop on the given point, we will get a similar figure :

Figure: The slope of the graph at given points

02

 Velocity graph

a. Hence we can try to plot the graph using this given data :

(b)

The time at which instantaneous velocity is greatest is td as the maximum value of the velocity is seen at time td.

(c)

Using the above slops in the graph, we can plot the value of velocity. When the slop is horizontal, velocity is ZERO.

When the slop is vertical, the slope increases to infinite hence velocity will also be infinite.

Instantaneous velocity is zero at tc, te, tg, tl because, at this point, the graph is touching the x-axis that, is the value of the velocity is becoming zero at this point.

(d)

The velocity is negative at ta, tb, and tf. because at this point, the graph has a negative value of the velocity

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Most popular questions from this chapter

Give an example in which there are clear distinctions among distance traveled, displacement, and magnitude of displacement. Specifically, identify each quantity in your example

A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate its velocity just before it strikes the floor. (b) Calculate its velocity just after it leaves the floor on its way back up. (c) Calculate its acceleration during contact with the floor if that contact lasts 0.0800 ms (\({\bf{8}}{\bf{.00 \times 1}}{{\bf{0}}^{{\bf{ - 5}}}}\;{\bf{s}}\)) . (d) How much did the ball compress during its collision with the floor, assuming the floor is absolutely rigid?

a) Explain how you can use the graph of position versus time in Figure 2.54 to describe the change in velocity over time.

Identify

(b) the time ( ta, tb , tc , td , or te ) at which the instantaneous velocity is greatest,

(c) the time at which it is zero, and

(d) the time at which it is negative.

Figure 2.54

a) Take the slope of the curve in Figure 2.64 to find the jogger’s velocity at t = 2.5 s. (b) Repeat at 7.5 s. These values must be consistent with the graph in Figure 2.65.

a) A light-rail commuter train accelerates at a rate of\({\bf{1}}{\bf{.35}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\). How long does it take to reach its top speed of 80.0 km/h, starting from rest? (b) The same train ordinarily decelerates at a rate of\({\bf{1}}{\bf{.65}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\). How long does it take to come to a stop from its top speed? (c) In emergencies the train can decelerate more rapidly, coming to rest from 80.0 km/h in 8.30 s. What is its emergency deceleration in\({\bf{m/}}{{\bf{s}}^{\bf{2}}}\)?

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