a body moves for a total of 9 seconds starting from rest with uniform acceleration and then with uniform retardation which is twice of value of acceleration to stop .what is the duration of uniform acceleration
Distance covered in 4 secods after tha 5th seconds= S9- s5 .tha distance covered in t second is given by s = ut 1/2 at2and distance covered in nth second is given by Dn = u a/2 (2n-1) The modern sr. sec. Bagpur palwal phone no. 8059892363
Use v=u+at and form two equations one for acc. And one for . Then on solving u will get d ans. I hope dat helps.. 😀 ☺
5s
Yesss
To find the duration of uniform acceleration, we first need to understand the problem. Let's break it down step by step.
1. Start from rest: The body starts from rest, which means its initial velocity is zero.
2. Uniform acceleration: The body accelerates uniformly for a certain period of time.
3. Uniform retardation: After the period of uniform acceleration, the body undergoes uniform retardation to come to a stop. The value of this retardation is twice the value of the acceleration during uniform acceleration.
4. Total duration: The total duration of the motion is given as 9 seconds.
Now, let's find the duration of uniform acceleration.
Let's assume the duration of uniform acceleration is 't' seconds.
During uniform acceleration:
- Initial velocity (u) = 0
- Final velocity (v) = ?
- Acceleration (a) = ?
- Time (t) = ?
From the given information, we know that the body accelerates uniformly for time 't'. Therefore, we can use the formula:
v = u + at
Since the initial velocity (u) is zero, the formula simplifies to:
v = at
Now, let's find the final velocity during uniform acceleration. We need to know the value of acceleration (a), which is not given directly.
From the problem statement, we know that the acceleration during uniform retardation is twice the acceleration during uniform acceleration. Let this acceleration be 'a_acc' for uniform acceleration and 'a_ret' for uniform retardation.
Therefore, a_ret = 2 * a_acc
Now, we can use the equation of motion to find 'a_acc':
First equation of motion: v^2 = u^2 + 2as
For uniform acceleration, initial velocity (u) is zero, and final velocity (v_acc) can be written as 'v_acc = a_acc * t' (since the body started from rest and we are calculating for uniform acceleration time).
The equation becomes:
(a_acc * t)^2 = 0 + 2 * a_acc * s
Simplifying, we get:
(a_acc^2 * t^2) = 2 * a_acc * s
Canceling 'a_acc' from both sides:
a_acc * t^2 = 2s
Finally, rearranging the equation, we get:
t = sqrt(2s / a_acc)
Since the retardation is twice the value of acceleration during uniform acceleration, we can substitute 'a_ret' for '2a_acc':
t = sqrt(2s / a_ret)
Now, we have the expression for the duration of uniform acceleration in terms of retardation.
We are given that the total duration of motion is 9 seconds. So, we can set up an equation using this information:
t (uniform acceleration) + t (uniform retardation) = 9
Substituting the expression for 't' in terms of retardation, we get:
sqrt(2s / a_ret) + sqrt(2s / a_ret) = 9
Simplifying this equation will give the value of 't', which is the duration of uniform acceleration. However, we require the values of 's' (displacement) and 'a_ret' (uniform retardation) to solve this equation. If you provide those values, I can help you find the duration of uniform acceleration.