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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

Short Answer

Expert verified

a) In the distance versus time graph, velocity is determined by the slope of the graph.

b) At time ta, the instantaneous velocity is maximum.

c) The time at velocity is zero is td.

d) The time at which velocity is decreasing is te

Step by step solution

01

Determine the velocity using slope

The velocity is the ratio of the displacement traveled to the time taken to travel that displacement.

If we draw a graph of displacement and time, we get a curve,

At any point, if we take the slope of the graph, it will give the value of velocity.

(a) Here, we can see that at point a, the slope of tangent to point a is maximum.

Hence it can be said that the value ofvelocity is maximum at A.

As we go from a to b, the slope value decreases continuously.

Hence by watching the slope of the graph, it can be said that whether the velocity will increase or decrease.

Slop is decreasing

So if we plot a graph of velocity versus time, we can answer the remaining questions.

V = slope

V = tanθ

In addition, the tangent's angle on the graph is dropping. As a result, the slope will decrease. It indicates that the particle's velocity is decreasing.

02

Velocity versus time graph

Velocity versus time graph

b)It can be seen from the accompanying graph of instantaneous velocity vs. time that thevelocity is maximum at ta.

c)It can be seen from the preceding graph of instantaneous velocity vs. time that thevelocity is 0 at the td.

d)It can be seen from the preceding graph of instantaneous velocity vs. time that thevelocity is negative at that te.

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Most popular questions from this chapter

Find the following for path C in Figure 2.59:

(a) The distance travelled.

(b) The magnitude of the displacement from start to finish.

(c) The displacement from start to finish.

(a) By taking the slope of the curve in Figure 2.60, verify that the velocity of the jet car is\({\bf{115}}{\rm{ }}{\bf{m}}/{\bf{s}}\)at\(t = {\rm{ }}{\bf{20}}{\rm{ }}{\bf{s}}\). (b) By taking the slope of the curve at any point in Figure 2.61, verify that the jet car’s acceleration is\({\bf{5}}.{\bf{0}}{\rm{ }}{\bf{m}}/{{\bf{s}}^{\bf{2}}}\).

A commuter backs her car out of her garage with an acceleration of\({\bf{1}}{\bf{.40}}\;{\bf{m/}}{{\bf{s}}^{\bf{2}}}\). (a) How long does it take her to reach a speed of 2.00 m/s? (b) If she then brakes to a stop in 0.800 s, what is her deceleration?

Dragsters can actually reach a top speed ofin onlyconsiderably less time than given in Example 2.10 and Example 2.11.

(a) Calculate the average acceleration for such a dragster.

(b) Find the final velocity of this dragster starting from rest and accelerating at the rate found in (a) for 402 m(a quarter mile) without using any information on time.

(c) Why is the final velocity greater than that used to find the average acceleration?

Hint: Consider whether the assumption of constant acceleration is valid for a dragster. If not, discuss whether the acceleration would be greater at the beginning or end of the run and what effect that would have on the final velocity.

In World War II, there were several reported cases of airmen who jumped from their flaming airplanes with no parachute to escape certain death. Some fell about 20,000 feet (6000m), and some of them survived, with few life-threatening injuries. For these lucky pilots, the tree branches and snow drifts on the ground allowed their deceleration to be relatively small. If we assume that a pilot’s speed upon impact was 123 mph (54 m/s), then what was his deceleration? Assume that the trees and snow stopped him over a distance of3.0m.

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