An object with initial velocity of 20 m/s accelerates uniformly at 5 m/s in the direction of its motion for a distance of 10 m. What is the final velocity of the object?
a = 5
v = 5t + c
when t = 0, v = 20
20 = 0+c ---> c = 20
then
v = 5t + 20
when s is the distance,
s = (5/2)t^2 + 20t + k
when t = 0, s = 0
0 = 0 + 0 + k ---> k = 0
s = (5/2)t^2 + 20t
how long did it take to go 10 m ?
10 = (5/2)t^2 + 20t
20 = 5t^2 + 40t
t^2 + 8t = 5
I will complete the square, which is the easiest way for this equation
t^2 + 8t + 16 = 5 + 16
(t+4)^2 = 21
t+4 = √21
t = √21-4 seconds
for that value of t,
v = 5(√21 - 4) + 20 m/s or appr 22.9 m/s
To find the final velocity of the object, we can use the following formula:
v^2 = u^2 + 2as
Where:
v = final velocity
u = initial velocity
a = acceleration
s = distance
Let's substitute the given values into the equation:
v^2 = (20 m/s)^2 + 2 * 5 m/s^2 * 10 m
v^2 = 400 m^2/s^2 + 100 m^2/s^2
v^2 = 500 m^2/s^2
Now, take the square root of both sides of the equation to solve for v:
v = √500 m/s
v ≈ 22.36 m/s
Therefore, the final velocity of the object is approximately 22.36 m/s.
To find the final velocity of the object, we can use one of the equations of motion known as the equation of uniformly accelerated motion:
v^2 = u^2 + 2as
where:
v = final velocity
u = initial velocity
a = acceleration
s = distance
In this case, we are given:
u = 20 m/s (initial velocity)
a = 5 m/s² (acceleration)
s = 10 m (distance)
Plugging in these values into the equation, we can solve for v:
v^2 = (20 m/s)^2 + 2 * 5 m/s² * 10 m
v^2 = 400 m²/s² + 100 m²/s²
v^2 = 500 m²/s²
To find v, we can take the square root of both sides of the equation:
v = √(500 m²/s²)
v ≈ 22.36 m/s
Therefore, the final velocity of the object is approximately 22.36 m/s.