Question

Hello !
Pls help me in the 8th question .
This question starts from the below of the left hand side of the page.
Pls match your answer with the answer key.
NEED CONTENT QUALITY ANSWER.
DETAILED EXPLANATION.
SPAMMERS STAY AWAY.

Answer 1

The velocity-time graph here represents the variation of the velocity of race-horse with respect to time.

Since velocity is a vector quantity, direction matters. Positive and negative velocity both carry different meanings.

A race-horse runs on a straight track. So, if we consider the movement in one direction as positive, then the movement in the opposite direction is considered negative.

The area under the v-t graph gives displacement during a particular time interval.

As we mentioned earlier, the direction also matters.

Let us consider the direction away from the initial point as positive.

We see that the horse is moving away from the initial point till t=30 min.

After t=30 min, the velocity becomes negative. This means that now the horse starts moving back towards the initial point.

A nice shaded graph is attached for a much-better visual understanding.

___________________________

Let us first Calculate the Areas of each of the Shaded Parts

$\displaystyle \textbf{ABC} = \frac{1}{2} \times \left(60 \, \, \frac{m}{min}\right) \times ((10-0) \, \, min) = \bold{300 \, \, m}$

$\displaystyle \textbf{BCED} = \left(60 \, \, \frac{m}{min}\right) \times ((15-10) \, \, min) = \bold{300 \, \, m}$

$\displaystyle \textbf{EFD} = \frac{1}{2} \times \left(60 \, \, \frac{m}{min}\right) \times ((30-15) \, \, min) = \bold{450 \, \, m}$

$\displaystyle \textbf{FGH} = \frac{1}{2} \times \left(40 \, \, \frac{m}{min}\right) \times ((40-30) \, \, min) = \bold{200 \, \, m}$

$\displaystyle \textbf{GHI} = \frac{1}{2} \times \left(40 \, \, \frac{m}{min}\right) \times ((55-40) \, \, min) = \bold{300 \, \, m}$

Now, we can solve all the questions easily.

______________________________

(a) Total Distance travelled by the Horse

To find the total Distance, we consider the positive area of each shaded region.

So,

$d=\text{ABC+BCED+EFD+FGH+GHI}\\ \\ \\ \implies d = 300+300+450+200+300 \\ \\ \\ \implies \boxed{\bold{d=1550 \, \, m}}$

Thus, Total Distance Travelled by Horse is 1550 metres.

(b) Displacement of Horse till 40 minutes

Here, we are asked Displacement. So the direction of velocity matters. Since negative velocity means that the horse is coming back to the initial point, areas under the graph on the negative side of the Y-Axis indicate negative displacement.

So, the areas on the negative side of Y-Axis are to be considered negative.

So, if Displacement till 40 minutes is s, then:

$s=\text{ABC+BCED+EFD-FGH}\\\\\\\implies s=300+300+450-200 \\ \\ \\ \implies \boxed{\bold{s=850 \, \, m}}$

Thus, The Displacement of the Horse from initial point till 40 minutes is 850 metres.

(c) Average Velocity of the Horse when it travels away from the initial point

Average Velocity is defined as the Total Displacement divides by the Total Time.

So,

$v_{avg}=\frac{\text{Displacement}}{\text{Time}}$

Now, we are asked to find the average velocity when the horse is moving away from the initial point.

This simply means that we do not have to consider the time when the horse turns back (velocity becomes negative) and starts to come towards the initial point.

The velocity becomes negative at t=30 min.

Thus, the horse is moving away from the initial point only till t=30 min.

So, we have to find the average velocity in the time interval of t=0 to t=35 min.

Total Displacement in this time interval (say s') is:

$s'=\text{ABC+BCED+EFD}\\\\\\\implies s'=300+300+450\\\\\\\implies \bold{s'=1050 \, \, m}$

So, Average Velocity would become:

$v_{avg} = \frac{\text{Displacement}}{\text{Time}} \\ \\ \\ \implies v_{avg}=\frac{s'}{t}\\\\\\\implies v_{avg}=\frac{1050 \, \, m}{30 \, \, m} \\ \\ \\ \implies \boxed{\bold{v_{avg}=35 \, \, m/min}}$

Thus, The Average Velocity of the Horse in the given time interval is 35 m/min.

[As you would have noticed, there is a correction in the unit of the answer in the original question]

Answer 2

Answer:

The velocity-time graph here represents the variation of the velocity of race-horse with respect to time.

Since velocity is a vector quantity, direction matters. Positive and negative velocity both carry different meanings.

A race-horse runs on a straight track. So, if we consider the movement in one direction as positive, then the movement in the opposite direction is considered negative.

The area under the v-t graph gives displacement during a particular time interval.

As we mentioned earlier, the direction also matters.

Let us consider the direction away from the initial point as positive.

We see that the horse is moving away from the initial point till t=30 min.

After t=30 min, the velocity becomes negative. This means that now the horse starts moving back towards the initial point.

A nice shaded graph is attached for a much-better visual understanding.

___________________________

Let us first Calculate the Areas of each of the Shaded Parts

\displaystyle \textbf{ABC} = \frac{1}{2} \times \left(60 \, \, \frac{m}{min}\right) \times ((10-0) \, \, min) = \bold{300 \, \, m}ABC=

2

1

×(60

min

m

)×((10−0)min)=300m

\displaystyle \textbf{BCED} = \left(60 \, \, \frac{m}{min}\right) \times ((15-10) \, \, min) = \bold{300 \, \, m}BCED=(60

min

m

)×((15−10)min)=300m

\displaystyle \textbf{EFD} = \frac{1}{2} \times \left(60 \, \, \frac{m}{min}\right) \times ((30-15) \, \, min) = \bold{450 \, \, m}EFD=

2

1

×(60

min

m

)×((30−15)min)=450m

\displaystyle \textbf{FGH} = \frac{1}{2} \times \left(40 \, \, \frac{m}{min}\right) \times ((40-30) \, \, min) = \bold{200 \, \, m}FGH=

2

1

×(40

min

m

)×((40−30)min)=200m

\displaystyle \textbf{GHI} = \frac{1}{2} \times \left(40 \, \, \frac{m}{min}\right) \times ((55-40) \, \, min) = \bold{300 \, \, m}GHI=

2

1

×(40

min

m

)×((55−40)min)=300m

Now, we can solve all the questions easily.

______________________________

(a) Total Distance travelled by the Horse

To find the total Distance, we consider the positive area of each shaded region.

So,

\begin{lgathered}d=\text{ABC+BCED+EFD+FGH+GHI}\\ \\ \\ \implies d = 300+300+450+200+300 \\ \\ \\ \implies \boxed{\bold{d=1550 \, \, m}}\end{lgathered}

d=ABC+BCED+EFD+FGH+GHI

⟹d=300+300+450+200+300

⟹

d=1550m

Thus, Total Distance Travelled by Horse is 1550 metres.

(b) Displacement of Horse till 40 minutes

Here, we are asked Displacement. So the direction of velocity matters. Since negative velocity means that the horse is coming back to the initial point, areas under the graph on the negative side of the Y-Axis indicate negative displacement.

So, the areas on the negative side of Y-Axis are to be considered negative.

So, if Displacement till 40 minutes is s, then:

\begin{lgathered}s=\text{ABC+BCED+EFD-FGH}\\\\\\\implies s=300+300+450-200 \\ \\ \\ \implies \boxed{\bold{s=850 \, \, m}}\end{lgathered}

s=ABC+BCED+EFD-FGH

⟹s=300+300+450−200

⟹

s=850m

Thus, The Displacement of the Horse from initial point till 40 minutes is 850 metres.

(c) Average Velocity of the Horse when it travels away from the initial point

Average Velocity is defined as the Total Displacement divides by the Total Time.

So,

v_{avg}=\frac{\text{Displacement}}{\text{Time}}v

avg

=

Time

Displacement

Now, we are asked to find the average velocity when the horse is moving away from the initial point.

This simply means that we do not have to consider the time when the horse turns back (velocity becomes negative) and starts to come towards the initial point.

The velocity becomes negative at t=30 min.

Thus, the horse is moving away from the initial point only till t=30 min.

So, we have to find the average velocity in the time interval of t=0 to t=35 min.

Total Displacement in this time interval (say s') is:

\begin{lgathered}s'=\text{ABC+BCED+EFD}\\\\\\\implies s'=300+300+450\\\\\\\implies \bold{s'=1050 \, \, m}\end{lgathered}

s

′

=ABC+BCED+EFD

⟹s

′

=300+300+450

⟹s

′

=1050m

So, Average Velocity would become:

\begin{lgathered}v_{avg} = \frac{\text{Displacement}}{\text{Time}} \\ \\ \\ \implies v_{avg}=\frac{s'}{t}\\\\\\\implies v_{avg}=\frac{1050 \, \, m}{30 \, \, m} \\ \\ \\ \implies \boxed{\bold{v_{avg}=35 \, \, m/min}}\end{lgathered}

v

avg

=

Time

Displacement

⟹v

avg

=

t

s

′

⟹v

avg

=

30m

1050m

⟹

v

avg

=35m/min

Thus, The Average Velocity of the Horse in the given time interval is 35 m/min.

[As you would have noticed, there is a correction in the unit of the answer in the original question]