Question

The graph shows the distance of a car from a measuring position located on the edge of a straight road.(a) What was the average velocity of the car from t = 0 to t = 30 seconds?(b) What was the average velocity of the car from t = 10 to t = 30 seconds?(c) About how fast was the car traveling att = 10 seconds? at t = 20 s ? at t = 30 s ? (d) What does the horizontal part of the graph between t = 15 and t = 20 seconds mean?(e) What does the negative velocity at t = 25 represent?

Answer 1

Answer:

The figure is missing: find it in attachment.

(a)

The average velocity of an object is given by

$v=\frac{d}{t}$

where:

d is the displacement (the change in position)

t is the time elapsed

Here we want to know the average velocity of the car between t = 0 and t = 30 s, so the time interval is

$t=30-0 = 30 s$

We notice that:

- at t = 0, the distance from the origin is x = 0

- at t = 30 s, the distance from the origin is x = 300 ft

So, the displacement is:

$d=300-0 = 300 ft$

Therefore, the average velocity in this part is:

$v=\frac{300}{30}=10 ft/s$

(b)

Here we want to find the average velocity from t = 10 to t = 30 seconds.

Given the two points

t = 10 s

t = 30 s

The time elapsed in this case is

$t=30-10 = 20 s$

On the other hand, we notice that:

- At t = 10 s, the distance from the origin is x = 100 ft

- At t = 30 s, the distance from the origin is x = 300 ft

Therefore, the displacement is

$d=300-100 = 200 ft$

And so, the average velocity of the car in this phase is

$v=\frac{200}{20}=10 ft/s$

(c)

Here we are asked to find how fast is the car going at some precise instants in time: this means, we are asked to find the instantaneous velocity of the car at a certain moments.

This can be found by evaluating the slope of the curve at those precise instants of time.

At t = 10 s, we can estimate the slope by looking at the 5-seconds interval before and after t = 10. We see that in this interval, the distance covered increases by approx 150 ft (from 50 to 200) in a time interval of 10 s, so the instantaneous velocity is approximately

$v_{10}=\frac{150}{10}=15 ft/s$

At t = 20 s, the situation is easier: in fact, we see that the curve is horizontal, this means that the slope is zero, so the instantaneous velocity is zero:

$v_{20}=0$

Finallly, at t = 30 s, we evaluate the  slope by looking at the few seconds interval before. We see that the distance covered increases by 100 ft in approx. 3 seconds, so the instantaneous velocity at t = 30 s is

$v_{30}=\frac{100}{3}=33.3 ft/s$

(d)

In a distance-time graph, the slope of the graph is equal to the velocity of the graph.

In fact, velocity is the ratio between displacement and time interval

$v=\frac{d}{t}$

The slope of a graph is given by the ratio between the change in the y-variable and the change in the x-variable:

$m=\frac{\Delta y}{\Delta x}$

However for this graph, $\Delta y$ corresponds to the displacement, while $\Delta x$ corresponds to the time interval, therefore the slope corresponds to the velocity.

In this graph, the curve between t = 15 s and t = 20 s is horizontal: this means that the slope is zero. Therefore, the velocity of the car in that interval is also zero.

(e)

As we said before, the velocity of the car is the ratio between the displacement and the time interval:

$v=\frac{d}{t}$

At t = 25 s, the velocity of the car is negative. Looking at the equation, this means that the displacement is also negative. This means that the distance from the origin at a later time is smaller than the distance from the origin at a earlier time.

If we interpret this, it basically means that the car is moving towards the origin, instead that away from it: therefore,

The negative velocity at t = 25 s means that at t = 25 s the car is moving backward, towards the origin.

Answer 2

At time t=25s the graph is showing that the distance between the current location of the car and the origin is reducing.

What is the relation between speed, distance, and Time?

We know that speed, distance, and time all are in a relationship to each other. this relationship can be given as,

$\rm{Speed = \dfrac{Distance}{Time}$

A.) We know that to find out the average velocity between time period t =0 to t=30, we need to find the ratio of the distance and time between time period t =0 to t=30.

$\rm Average\ Velocity =\dfrac{\text{Total Distance traveled by the car}}{\text{Total time taken by the car}}$

$\begin{aligned}\rm Average\ Velocity &=\dfrac{D_{30}-D_0}{T_{30}-T_0}\\\\& =\dfrac{300-0}{30-0}\\\\& = 10\ ft/sec \end{aligned}$

B.) We know that to find out the average velocity between time period t =10 to t=30, we need to find the ratio of the distance and time between time period t =10 to t=30.

$\rm Average\ Velocity =\dfrac{\text{Total Distance traveled by the car}}{\text{Total time taken by the car}}$

$\begin{aligned}\rm Average\ Velocity &=\dfrac{D_{30}-D_{10}}{T_{30}-T_{10}}\\\\& =\dfrac{300-100}{30-10}\\\\& = 10\ ft/sec \end{aligned}$

C.) To find the velocity of the car, simply find the slope of the line at that point in time.

At t=10,

If we try to find the slope of the function at t=10,

$m= \dfrac{\triangle y}{\triangle x} = \dfrac{200-50}{15-5} = 15\rm\ ft/sec$

At t=20,

If we try to find the slope of the function at t=20,

$m= \dfrac{\triangle y}{\triangle x} = \dfrac{200-200}{20-15} = 0\rm\ ft/sec$

At t=30,

If we try to find the slope of the function at t=30,

$m= \dfrac{\triangle y}{\triangle x} = \dfrac{300-100}{30-27.5} = 80\rm\ ft/sec$

D.) The horizontal part between the time period 15s to 20s shows that there is no moment in the car, therefore, the car is at rest. because we closely look at the graph, we can observe that the time is increasing but the distance of the car is not moving.

Thus, the car is at rest between the time period 15s to 20s.

E.) The negative velocity at t=25 shows that the body is moving towards the origin(starting point), as the distance between the starting point and the is reducing, therefore, the graph is showing that the distance between the current location of the car and the origin is reducing.

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