Question

In heavy rush-hour traffic you drive in a straight line at 12 m/s for 1.5 milrutes, then you have to stop for 3.5 minutes, and finally i you drive at 15 m/s for another 2.5 minutes. (a) Plot a position-versus-time graph for this motion. Your plot should extend from t= 0 to t =7 .5 minutes. (b) Use your plot from part (a) to calculate the average velocity between t= 0 and t =
7.5 minutes.

Answer 1

Average velocity  between t= 0 and t = 7.5 minutes = 7.4 m / s

Further explanation

Regular straight motion is the motion of objects on a straight track that has a fixed speed

Formula used

$\large{\boxed{\boxed{\bold{S\:=\:v\:\times\:t}}}$

S = distance = m

v = speed = m / s

t = time = seconds

The motion of objects can be expressed in graphical form

This relationship graph can be in the form of a graph S-t, v-t or a-t

From the S-t graph, the average speed can be determined by determining the slope distance of the curve line from the distance to time ratio

average velocity = distance traveled / time taken

 

From the statement of questions there are 3 stages of time that occur

• 1. first time

move with v = 12 m / s, t = 1.5 min = 90 s

So the distance traveled =

s = v x t

S = 12. 90

S = 1080 m

• 2. second time

stop (v = 0) so that s = 0 and t = 3.5 min = 210 s

• 3. third time

move with v = 15 m / s and t = 2.5 min = 150 s

so the distance traveled

s = v x t

s = 15 x 150

s = 2250 m

Total distance = 2250 + 1080 = 3330  m

If we look at the existing chart (attached)

then we can draw a straight line from the starting point to the end point so we get a slash curve that shows the average speed

v average = distance traveled / time taken

v average = 3330/450

v average = 7.4 m / s

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Answer 2

(a)The plot of the position versus time graph is shown in the figure below.

(b)The average velocity of the car in the interval $0 - 7.5\,\min$ is $\boxed{7.4\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$.

Further Explanation:

Part (a):

The velocity of a body is defined as the rate of change of the position of a body.

The distance covered by the car in first $1.5\,\min$ or $90\,\sec$ is given as:

${d_1} = {v_1} \times t$  

Substitute the values in equation.

$\begin{aligned}{d_1} &= 12 \times 90\\&= 1080\,{\text{m}}\\\end{aligned}$  

The distance covered by the car in the last $2.5\,\min$ or $150\,\sec$ is given as:

${d_2} = {v_2} \times t$  

Substitute the values in equation.

$\begin{aligned}{d_2} &= 15 \times 150 \\&= 2250\,{\text{m}} \\\end{aligned}$  

Thus, the car starts from origin and reaches the $1080\,{\text{m}}$ distance in first $1.5\,\min$ and then it remains at rest for next $3.5\,\min$ and after that it again starts moving and covers $2250\,{\text{m}}$ further in next $2.5\,\min$. The plot for the position of car versus time is as shown below.

Part (b):

The average velocity of the car is obtained as the overall rate of the distance covered by the car. The average velocity of the car is obtained as:

${v_{avg}} = \dfrac{{{x_f} - {x_i}}}{t}$  

Here, ${x_f}$ is the final position and ${x_i}$ is the initial position of the car.

Substitute the values in above expression.

$\begin{aligned}{v_{avg}} &= \frac{{\left( {1080 + 2250} \right) - 0}}{{\left( {7.5 \times 60} \right)}} \\&=\frac{{3330}}{{450}}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&= 7.4\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$  

Thus, the average velocity of the car during the whole journey is $\boxed{7.4\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}$.

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Answer Details:

Grade: Senior School

Chapter: Speed and velocity

Subject: Physics

Keywords:  Heavy rush-hour, straight line, 12 m/s, 1.5 min, speed of car, time taken, average velocity, rate of change of positon, position versus time, slope.