a particle at rest starts in horizontal strsight line with uniform acceleration ratio distance during 4th and 3rd sec
ond
distance=1/2 a t^2
distance during third second:
1/2 a (3^2-2^2)
distance during fourth second:
1/2 a (4^2-3^2)
ratio distance during 4th to 3rd:
(16-9)/(9-4)=7/5
To find the acceleration of the particle, we need to use the kinematic equation for displacement. The equation is:
\(d = ut + \frac{1}{2}at^2\)
Where:
- d is the displacement
- u is the initial velocity (which is zero since the particle starts at rest)
- a is the acceleration
- t is the time
In this case, we know that the particle starts at rest, so u = 0. We also know that the time interval between the fourth second and the third second is 1 second (4th sec - 3rd sec = 1 sec). Therefore, we can calculate the displacement during this time interval.
Let's denote the displacement during the fourth second as \(d_4\) and during the third second as \(d_3\). Using the equation above, we can write:
\(d_4 = 0 + \frac{1}{2}a(4^2)\)
\(d_3 = 0 + \frac{1}{2}a(3^2)\)
From the given information, we can see that the distance traveled during the fourth second is twice the distance traveled during the third second. Therefore, we can write the equation:
\(2d_3 = d_4\)
Now, let's substitute the values of \(d_4\) and \(d_3\) into this equation:
\(2(\frac{1}{2}a(3^2)) = \frac{1}{2}a(4^2)\)
Simplifying, we get:
\(2(9a) = 16a\)
\(18a = 16a\)
\(2a = 0\)
From this equation, we can see that the acceleration (a) is equal to zero. Therefore, the particle is not undergoing any acceleration during this time interval.