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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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 x 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 x 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
64 364 × S
28 = 4 x 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