Question

A car starts with constant acceleration a = 2m/s² at t = 0. Two coins are released from the car at t = 3 & t = 4. Each coin takes 1 second to fall on ground. Then the distance between the two coins will be (Assume coin sticks to the ground)



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Answer 1

Given :-

• The acceleration of car (a) = 2m/s²
• Time taken by 1st coin (t_1) = 3s
• Time taken by 2nd coin (t_2) = 4s

To Find :-

• The distance between two coins = ?

Solution :-

• To calculate distance between two coins at first have find the final velocity of both coins then calculate it's distance.

Calculation for 1st coins :-

• As per 1st equation of motion :-]

⇢ V = U + AT

• Initial velocity (u) = 0
• Final velocity (v) = ?
• Time taken (T) = 3s

⇢ V = 0 + 2 × 3

⇢ V = 0 + 6

⇢ V = 6m/s

Calculation for 2nd coins :-

• As per 1st equation of motion :-]

⇢ V = U + AT

• Initial velocity (u) = 0
• Final velocity (v) = ?
• Time taken (T) = 4s

⇢ V = 0 + 2 × 4

⇢ V = 0 + 8

⇢ V = 8m/s

Now calculate distance between two coins :-

• In the given Question The two coins take 1s to fall on the ground at vertically on the ground.

By using 3rd equation of motion :-

⇢ V² - U² = 2AS

• Final velocity (v) = 8m/s
• Initial velocity (u) = 6m/s
• Acceleration (a) = 2m/s²

⇢ 8² - 6² = 2 × 2 × S

⇢ 64 - 36 = 4 × S

⇢ 28 = 4 × S

⇢ S = 28/4

⇢ S = 7m

• Notice here the car is releasing two coins at different times 3s and 4s but both the two coin is falling on the ground at the same time. It's mean here time taken to cover distance between the two coin is 1s the distance covered by coin in 1s equal to 1m because the difference between its time is 1s.

Distance between two coins :-

⇢ Distance = Distance covered by car between the two coin + Distance covered coins in 1s

⇢ Distance = 7m + 1m + 1m

⇢ Distance = 9m

Hence, The required distance = 9m :-

Answer 2

Information provided with us:

• A car starts with constant acceleration a = 2m/s at t = 0
• Two coins are released from the car at t = 3 and t = 4.
• Each coin takes 1 second to fall on ground

Assumptions already been taken:

• Coins had been sticked to the ground

What we have find out:

• We have find out the distance between the two coins

Using Formulas:

First equation of motion:-

• v = u + at

Where,

• v is final velocity
• u is initial velocity
• t is time

Third equation of motion:-

• v² = u² + 2as

Where,

• v is final velocity
• u is initial velocity
• t is time

We have,

• time taken by the first coin is 3s
• time taken by the second coin is 4s
• Initial velocity is always 0

Required Solution:

Finding out the final velocity of the first coin:

Substituting the values in first equation of motion,

• We know that intial velocity is always zero

➻ v = 0 + 2(3)

➻ v = 0 + 2×3

➻ v = 0 + 6

➻ v = 6

• Therefore, final velocity (v) of the first coin is of 6 ms-¹

Finding out the final velocity of second coin:

Again substituting the values in first equation of motion,

• We know that intial velocity is always zero and time taken would be 4s

➻ v = 0 + 2(4)

➻ v = 0 + 2×4

➻ v = 0 + 8

➻ v = 8

• Therefore, final velocity of the second coin is 8ms-¹

Finding out distance:

Substituting the values in third equation of motion in order to calculate the distance,

• Here remember that Intial velocity would be 6 ms-¹ because it came to rest.

➺ 8² = 6² - 2(2)(s)

➺ 8² = 6² - 2×2(s)

➺ 8² = 6² - 4s

➺ 8×8 = 6² - 4s

➺ 64 = 6² - 4s

➺ 64 = 6×6 - 4s

➺ 64 = 36 - 4s

➺ 4s = 64 - 36

➺ 4s = 28

➺ s = 28/4

➺ s = 14/2

➺ s = 7

• Therefore, distance is of 7m

At last finding out distance:

In order to find out distance we would add the distance covered by both the coins with 7m. We know that a body travels 1m in 1s so both the coins would take 1m each respectively. As their are two coins so it would be 2s.

➺ D = 7m + 2m

➺ D = 9m

• Therefore, the distance between the two coins will be of 9m.

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