Question

${ \huge{\boxed{\sf{ \red{Question :-}}}}}$
Two green anacondas X and Y of length 10 m are moving in the same direction with a uniform speed of 20 kmph, with X ahead of Y. Upon detecting prey, Y decides to overtake X. Thus it accelerates by 1 m/s²,  after 50 s the tail of Y  slithers past  X’s head.  Calculate the original distance between the two?

✔ Need well explained answer.
✔ Wrong answer will be deleted.

✖ Don't Spam. ​

Answer 1

${\sf{\large{\bigstar{ \: Given}}}}$

For anaconda X

• Initial velocity, u =20kmph = 5.55 m/s
• Time, t = 50 secs.
• Acceleration , a = 0 (Because it is moving with a uniform speed)

For anaconda Y

• Initial velocity, u =20kmph = 5.55 m/s
• Time, t = 50 secs.
• Acceleration , a = 1 m/s²

${ \tiny{ \sf[* Converting \: km/h \: to \: m/s ]}}$

$\: \: \: {\sf{\tiny{ \implies\frac{1000}{3600} \times 20}}} \\ {\sf{\tiny{ \boxed{ \implies5.55 \: m /s}}}}$

${\sf{\large{\bigstar{ \: To \: find : }}}}$

• Original distance between them.

${\large{\sf{\bigstar{ \: Distance \: traveled \: by \: Anaconda \: X \: is }}}}$

Using the second equation of motion we get :

${\color{green}{ \boxed{\sf{S = ut+ \frac{1}{2} \: a(t)²}}}}$

${ \implies{\sf{S_X = (5.55)(50)+ \frac{1}{2} \: (0)(50)²}}}$

$\implies{\sf{S_X = 277.5+ 0 \times 2500}}$

$\implies{\sf{S_X = 277.5+ 0}}$

${ \boxed{ \implies{\sf{S_X = 277.5 \: m}}}}$

${\large{\sf{\bigstar{ \: Distance \: traveled \: by \: Anaconda \: Y \: is }}}}$

Using the second equation of motion we get :

${\color{green}{ \boxed{\sf{S = ut+ \frac{1}{2} \: a(t)²}}}}$

${ \implies{\sf{S_Y = (5.55)(50)+ \frac{1}{2} \: (1)(50)²}}}$

$\implies{\sf{S_Y = 277.5+ \frac{1}{2} \times 2500}}$

$\implies{\sf{S_Y = 277.5+ 1250}}$

${ \boxed{ \implies{\sf{S_Y = 1527.5\: m}}}}$

The original distance between the two is :-

${ \large{ \mathtt{ \implies{ \large 1527.5 \: – \: 277.5}}}}$

${ \large{ \red{ \boxed{ \implies{\sf{1250 \: m}}}}}}$

So, original distance between the two is 1250m.

Answer 2

Answer :

›»› The original distance between the two anacondas are 1250 m.

Given :

• Two green anacondas X and Y of length 10 m are moving in the same direction with a uniform speed of 20 kmph, with X ahead of Y. Upon detecting prey, Y decides to overtake X. Thus it accelerates by 1 m/s², after 50 s the tail of Y  slithers past  X’s head$.$

To Calculate :

• The original distance between the two.

How to calculate?

Here in this question we have to calculate the original distance between the two anacondas. So, firstly we need to convert the initial velocity from km/hr to m/s, after that we will find the distance of the anaconda X, after that we will find the distance of anaconda Y, after all this we will calculate the original distance between the two anacondas.

To convert the initial velocity of anaconda we will divide the speed value by 3.6.

To find the distance of anaconda X, we can use second equation of motion to find the distance of anaconda X, which says s = ut + ½ at².

And also, we can use second equation of motion to find the distance of anaconda Y which says s = ut + ½ at².

Here,

• s is the Distance.
• u is the Initial velocity.
• t is the Time taken.
• a is the Acceleration

Units,

• The unit of Distance is m.
• The unit of Initial velocity is m/s.
• The unit of Time taken is second.
• The unit of Acceleration is m/s².

Calculation :

==> Initial velocity = 20 km/hr

==> Initial velocity = 20 ÷ 3.6

==> Initial velocity = 5.55 m/s².

$\:$

For anaconda X :

• Initial velocity of anaconda X = 5.55 m/s
• Time taken by anaconda X = 50 sec.
• Acceleration of anaconda X = 0 m/s² (moving with a uniform speed).
• Distance of anaconda X = ?

From second equation of motion

⇛ s = ut + ½ at²

⇛ s = 5.55 × 50 + ½ × 0 × (50)²

⇛ s = 5.55 × 50 + ½ × 0 × 2500

⇛ s = 5.55 × 50 + ½ × 0

⇛ s = 5.55 × 50 + 0

⇛ s = 5.55 × 50 + 0

⇛ s = 277.5 + 0

⇛ s = 277.5

$\:$

For anaconda Y :

• Initial velocity of anaconda Y = 5.55 m/s.
• Time taken by anaconda Y = 50 sec.
• Acceleration of anaconda Y = 1 m/s².
• Distance of anaconda Y = ?

From second equation of motion

⇛ s = ut + ½ × at²

⇛ s = 5.55 × 50 + ½ × 1 × (50)²

⇛ s = 5.55 × 50 + ½ × 1 × 2500

⇛ s = 277.5 + ½ × 1 × 2500

⇛ s = 277.5 + ½ × 2500

⇛ s = 277.5 + 1 × 1250

⇛ s = 277.5 + 1250

⇛ s = 1527.5

Now,

The original distance between the two = Distance of anaconda Y - Distance of anaconda X.

==> 1527 - 277.5

==> 1250 m

Hence, the original distance between the two anacondas are 1250 m.