An object is moving with uniform acceleration and it's velocity after 5 second is 25m/s after 8 second it is 34m/s find distance travelled by object in 12 second
v(t)=vo+at
25=vo+5a
34=vo+8a
subtract 1 from 2
9=3a
a=3m/s^
25=vo+15
vo=10m/s
v(12)=10+3*12= 46m/s
avg velocity=(10+46)/2=28m/s
distance=28m/s*12sec...
To find the distance traveled by the object in 12 seconds, we can use the formula for displacement with uniform acceleration:
\(s = ut + \frac{1}{2}at^2\)
Where:
- \(s\) is the displacement (distance traveled)
- \(u\) is the initial velocity (at \(t = 0\))
- \(a\) is the acceleration
- \(t\) is the time
Given:
- The velocity after 5 seconds is 25 m/s. This is the initial velocity at \(t = 5\): \(u = 25\,m/s\)
- The velocity after 8 seconds is 34 m/s. This is the velocity at \(t = 8\): \(v = 34\,m/s\)
- We want to find the displacement at \(t = 12\): \(s = ?\)
First, let's find the acceleration using the given information:
\(v = u + at\) ...(Equation 1)
Substituting the values for \(v\), \(u\), and \(t\), we have:
\(34 = 25 + a(8)\)
Solving this equation, we find:
\(a = \frac{34 - 25}{8} = \frac{9}{8} m/s^2\)
Now that we have the acceleration, we can find the initial displacement using Equation 1:
\(v = u + at\)
Substituting the values for \(v\), \(u\), \(a\), and \(t\):
\(25 = u + \frac{9}{8} (5)\)
Simplifying this equation, we find:
\(u = 20\,m/s\)
Now, let's find the displacement at \(t = 12\) using the formula:
\(s = ut + \frac{1}{2}at^2\)
Substituting the values for \(u\), \(a\), and \(t\):
\(s = 20(12) + \frac{1}{2}\left(\frac{9}{8}\right)(12)^2\)
Simplifying this equation, we find:
\(s = 240 + \frac{27}{2}(12) = 240 + 162 = 402\,m\)
Therefore, the distance traveled by the object in 12 seconds is \(402\,m\).
To find the distance traveled by the object in 12 seconds, we need to use the equations of motion.
The first equation of motion is:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
From the given information, we can see that the initial velocity (u) is not mentioned, so we need to find it.
Using the first equation of motion for the first 5 seconds:
25 m/s = u + a * 5 s.
Similarly, for the next 3 seconds:
34 m/s = u + a * 8 s.
By subtracting the equations, we can eliminate the variable 'u':
34 m/s - 25 m/s = a * 8 s - a * 5 s.
9 m/s = 3a * s.
Dividing both sides by 3s:
3 m/s² = a.
Now, we can substitute the value of 'a' back into one of the original equations to find 'u.'
Using the equation:
25 m/s = u + a * 5 s,
25 m/s = u + 3 m/s² * 5 s,
25 m/s = u + 15 m/s.
Subtracting 15 m/s from both sides, we get:
u = 10 m/s.
Now that we have both the initial velocity (u) and acceleration (a), we can calculate the distance traveled in 12 seconds using the second equation of motion:
s = ut + (1/2) * a * t²,
where s is the distance traveled.
Plugging in the values:
s = 10 m/s * 12 s + (1/2) * 3 m/s² * (12 s)²,
s = 120 m + (1/2) * 3 m/s² * 144 s²,
s = 120 m + 216 m,
s = 336 m.
Therefore, the distance traveled by the object in 12 seconds is 336 meters.