Initial Velocity Formula: Initial Velocity, denoted as u, signifies the velocity at time interval t = 0, representing the velocity at the beginning of motion. There exist four formulas to calculate initial velocity:
When time (t), acceleration (a), and final velocity (v) are given, the initial velocity (u) is expressed as:
Initial Velocity Formula u=v−at
In the case of given values for final velocity (v), acceleration (a), and distance (s), the formula is:
Initial Velocity Formula u
2
=v
2
−2as
If the provided values are distance (s), acceleration (a), and time (t), the formula for initial velocity is:
Initial Velocity Formula u= s/t − 1/2 at
Where: Initial velocity is represented as u Final velocity is denoted as v Time taken is t Distance traveled or displacement is s Acceleration is a In the scenario where final velocity (v), distance (s), and time (t) are given, the formula for initial velocity is:
Initial Velocity Formula u= 2(s/t ) −v
Initial Velocity Formula Solved Examples
Example 1:
A car accelerates from rest at a rate of 4 m/s² for a duration of 5 seconds. Calculate the final velocity of the car.
Answer: Given: Initial velocity u = 0 m/s (as the car starts from rest) Acceleration a = 4 m/s² Time t = 5 s
Using the Initial Velocity Formula v=u+at
v=0+(4×5)
v=20m/s
Therefore, the final velocity v is 20 m/s.
Example 2:
An object is thrown vertically upward with an initial velocity of 15 m/s. If the acceleration due to gravity is 9.8 m/s 2 9.8m/s 2 , find the time it takes for the object to reach its highest point.
Answer: Given: Initial velocity u = 15 m/s Acceleration a (due to gravity) = −9.8m/s
2
(negative because it acts downward) Using the formula v=u+at, and knowing that at the highest point v=0
0=15−9.8t
Solving for t= 15/9.8
t≈1.53s
Therefore, it takes approximately 1.53 seconds for the object to reach its highest point.
Example 3:
A ball is dropped from a height of 25 meters. Determine the final velocity just before it hits the ground. Take the acceleration due to gravity as 9.8m/s
2
.
Answer: Given: Initial velocity u = 0 m/s (ball is dropped, so initial velocity is zero) Acceleration a (due to gravity) = 9.8m/s
2
Distance s (height) = 25 m
Using the formula v
2
=u
2
+2as
v
2
=0+2×9.8×25
v
2
=2×9.8×25
v
2
=490
Taking the square root to find v
2
= 490
v≈22.14m/s Therefore, the final velocity just before the ball hits the ground is approximately 22.14m/s.
Example 4:
A car decelerates from a speed of 30 m/s to rest in 5 seconds. Calculate the acceleration and the distance traveled during the deceleration.
Answer: Given: Initial velocity u = 30 m/s Final velocity v = 0 m/s (as the car comes to rest) Time t = 5 s
Using the formula a= v−u / t to find acceleration: a= 0−30/5
a=−6m/s 2
The negative sign indicates deceleration.
Using the formulas=ut+ 1/2 at 2 to find distance traveled:
s=30×5+ 2 1 ×(−6)×(5 2 )
s=150−75
s=75m
Therefore, the acceleration is −6m/s 2 (deceleration), and the distance traveled during the deceleration is 75 meters.
Example 5:
A car accelerates from rest at a rate of 3 m/s² for a duration of 5 seconds. Calculate the final velocity of the car.
Answer: Given: Initial velocity u = 0 m/s (as the car starts from rest) Acceleration a = 3 m/s² Time t = 5 s
Using the formula v=u+at
v=0+(3×5)
v=15m/s
Therefore, the final velocity v is 15 m/s.
Example 6:
A car accelerates from rest at a rate of 2 m/s² for a duration of 5 seconds. Calculate the final velocity of the car.
Answer: Given: Initial velocity u = 0 m/s (as the car starts from rest) Acceleration a = 2 m/s² Time t = 5 s
Using the formula v=u+at
v=0+(2×5)
v=10m/s
Therefore, the final velocity v is 10 m/s.
Example 7:
A car accelerates from rest at a rate of 8 m/s² for a duration of 5 seconds. Calculate the final velocity of the car.
Answer: Given: Initial velocity u = 0 m/s (as the car starts from rest) Acceleration a = 8 m/s² Time t = 5 s
Using the formula v=u+at
v=0+(8×5)
v=40m/s
Therefore, the final velocity v is 40 m/s.
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