A body moving in a straight line with an acceleration of 12ms-1 has a velocity of 8ms-1 after 5 seconds. Find the initial velocity of the body and the distance covered in the 9th second?
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What about the distance in the 9th second?
To find the initial velocity of the body, we can use the equation:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. We are given that the final velocity is 8 m/s, the acceleration is 12 m/s^2, and the time is 5 seconds. Plugging in these values, we get:
8 m/s = u + (12 m/s^2)(5 s).
Simplifying this equation, we have:
8 m/s = u + 60 m/s.
To isolate u, we can subtract 60 m/s from both sides of the equation:
u = 8 m/s - 60 m/s,
u = -52 m/s.
So, the initial velocity of the body is -52 m/s.
To find the distance covered in the 9th second, we need to find the velocity at the end of the 8th second. Since the acceleration is constant, we can use the equation:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. We know that the acceleration is 12 m/s^2 and the time is 8 seconds. Plugging in these values, we have:
v = -52 m/s + (12 m/s^2)(8 s).
Simplifying this equation, we get:
v = -52 m/s + 96 m/s.
v = 44 m/s.
Therefore, the velocity at the end of the 8th second is 44 m/s.
To find the distance covered in the 9th second, we can use the equation:
s = ut + 0.5at^2,
where s is the distance covered, u is the initial velocity, a is the acceleration, and t is the time. We know that the initial velocity is 44 m/s, the acceleration is 12 m/s^2, and the time is 1 second (since we want to find the distance covered in the 9th second). Plugging in these values, we have:
s = (44 m/s)(1 s) + 0.5(12 m/s^2)(1 s)^2,
s = 44 m + 0.5(12 m/s^2)(1 s)^2,
s = 44 m + 6 m,
s = 50 m.
Therefore, the distance covered in the 9th second is 50 meters.
To solve this problem, we can use the equations of motion.
Let's denote:
a = acceleration (given as 12 m/s^2)
v = final velocity (given as 8 m/s)
t = time (given as 5 seconds)
u = initial velocity (what we need to find)
t2 = time in the 9th second (what we need to find)
s = distance covered (what we need to find)
First, we can find the initial velocity (u):
We can use the equation:
v = u + at
Rearranging the equation to solve for u:
u = v - at
Substituting the given values:
u = 8 m/s - (12 m/s^2) * 5 s
u = 8 m/s - 60 m/s
u = -52 m/s
So, the initial velocity (u) of the body is -52 m/s (negative sign indicates that the body is moving in the opposite direction).
Next, let's find the distance covered in the 9th second (s2):
We can use the equation:
s = ut + (1/2)at^2
Since we need to find the distance covered in the 9th second, we'll use the time t2 = 9 seconds.
s2 = u * t2 + (1/2) * a * (t2)^2
Substituting the given values:
s2 = -52 m/s * 9 s + (1/2) * 12 m/s^2 * (9 s)^2
s2 = -468 m + (1/2) * 12 m/s^2 * 81 s^2
s2 = -468 m + 54 m/s^2 * 81 s^2
s2 = -468 m + 4374 m
s2 = 3906 m
So, the distance covered in the 9th second (s2) is 3906 meters.
I already went through this. What didn't you understand?
v(t) = v0 + at
v0 + 5*12 = 8
so v0 = -52 m/s
Now finish it off. How your work if you get stuck.