Question

A man runs 200m along a stràight track in the first 25s and then turns bàck instàntly and runs 100m in next 15s towards the straight point. The ratió of magnitude of average speed and average velôcity of the man is?

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Answer 1

Answer:

3:1

Explanation:

As per the provided information in the given question, we have:

• A man runs 200m along a straight track in the first 25s.
• Then, turns back instαntly and runs 100m in next 15s towards the straight point.

We have been asked to calculate the ratio of magnitude of average speed and average velocity of the man.

Before commencing the steps, have a look at the attachment. As per the question,

• From A to B, a man runs 200m along a straight track in the first 25s.
• Then, it turns back from B and travels to C that is, runs 100m in next 15s towards the straight point.

Now, let us denote the ratio of magnitude of average speed and average velocity as R. The ratio of magnitude of average speed and average velocity will be given by,

$\implies\boxed{\sf{\red{R=\dfrac{Speed_{av}}{Velocity_{av}}}}}$

Or,

$\implies\boxed{\sf{\red{R=Speed_{av}: Velocity_{av}}}}$

So, in order to calculate the required raito, we have to calculate the average velocity and average speed first.

Average Velocity :

Average velocity is calculated by dividing total displacement divided by total time. Mathematically,

$\bigstar\;\boxed{\sf{\red{Velocity_{av} = \dfrac{Total\; Displacement}{Total \; Time}}}}$

Displacement is the shortest distance from its initial position to the final position. Here, man's initial position is A and his final position is C. The shortest distance between any two points is always a straight line. Hence, AC is the displacement.

From the diagram,

$\implies\sf{AB= AC + BC}$

$\implies\sf{AC= AB - BC}$

• AB = 200 m
• BC = 100 m

$\implies\sf{AC= (200 - 100)\; m}$

$\implies\underline{\sf{Displacement \; (AC) = 100\; m}} \dots \Big \lgroup 1 \Big \rgroup$

Therefore, the displacement is 100 m towards the straight point.

And, total time is equal to the sum of time taken to go from A to B and from B to C.

$\implies\sf{Total \; Time = 25\;s + 15\;s}$

$\implies\underline{\sf{Total \; Time = 40 \;s}}\dots \Big \lgroup 2 \Big \rgroup$

Substitute the values from ( 1 ) and ( 2 ) in the formula of average velocity.

$\implies\sf{Velocity_{av} = \dfrac{100 \; m_{(towards\: the \:straight \:point)}}{40\; s}}$

$\implies\sf{Velocity_{av} = \dfrac{10 \; m_{(towards\: the \: straight \:point)}}{4\; s}}$

$\implies\boxed{\sf{Velocity_{av} = 2.5 \; ms^{-1}\; _{(towards\: the\: straight \:point)} } } \; \red{\bigstar}$

Therefore, average velocity of the man is 2.5 m/s towards the straight point.

Average Speed :

Average speed is calculated by dividing total distance by total time. Mathematically,

$\bigstar\;\boxed{\sf{\red{Speed_{av} = \dfrac{Total\; Distance}{Total \; Time}}}}$

As the body covers 200 m firstly and then 100 m. So, total distance will be 200 m + 100 m = 300 m. And, 2e have already calculated the total time [Equation 2] that is 40 seconds.

$\implies\sf{Speed_{av} = \dfrac{300 \; m}{40\; s}}$

$\implies\sf{Speed_{av} = \dfrac{30 \; m}{4\; s}}$

$\implies\boxed{\sf{Speed_{av} = 7.5 \; ms^{-1} }} \; \red{\bigstar}$

Therefore, average speed is 7.5 m/s.

Required Ratio :

Now, the ratio will be :

$\implies\boxed{\sf{\red{R=\dfrac{Speed_{av}}{Velocity_{av}}}}}$

$\implies\sf{R = \dfrac{7.5\; ms^{-1}}{2.5 \; ms^{-1}}}$

$\implies\sf{R = \dfrac{750}{250}}$

$\implies\sf{R = \dfrac{75}{25}}$

$\implies\sf{R = \dfrac{15}{5}}$

$\implies\sf{R = \dfrac{3}{1}}$

$\implies\underline{\boxed{\sf{\red{R = 3 : 1}}}}$

Therefore, the ratio of magnitude of average speed and average velocity of the man is 3:1.

$\rule{200}2$

Answer 2

Answer:

3:1

Explanation:

In this case, total distance covered = 200 m + 100 m = 300 m

Net displacement (∆x) = + 100 m

Total time = 25 + 15 s = 40 s

Avg speed = total distance/total time = 300/40 m/s

Avg speed = total distance/total time = 300/40 m/sAvg velocity = closest distance from final & initial point/total time = 100/40 m/s

Therefore,

ratio = 300/40 : 100/40 = 3:1.

More:

• Avg speed when particle has variable speed in diff time = Σviti/Σti
• Avg speed when a particle covers same distance is same time with variable speed => 1/v(av) = 1/nΣ1/vi.
• ∫a.dt = v, ∫v.dt = x, ∫x.dt = no physical significance.
• v = dx/dt, a = dv/dt & a = v dv/dx

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