a point mass starts moving in a straight line with a constant acceleration a .at a time t from beginning of motion ,the acceleration sign changes, without changing magnitude.determine the time t1from the beginning of motion in which the point of mass returns to initial point
To determine the time \( t_1 \) when the point mass returns to its initial position, we can use the kinematic equations.
Let's break down the problem step by step:
1. Initial conditions:
We can assume the initial position of the point mass is \( x_0 = 0 \) (start of motion) and the initial velocity is \( v_0 = 0 \) (point mass starts from rest).
2. Breaking the motion into two parts:
We can divide the motion into two phases: the initial phase with constant positive acceleration \( a \), and the later phase with constant negative acceleration \( -a \). The magnitude of acceleration remains the same throughout, but the direction changes.
3. Determining the time for each phase:
We can determine the time \( t_1 \) when the point mass returns to its initial position by calculating the time for each phase separately.
a) For the initial phase:
Using the equation \( x = x_0 + v_0t + \frac{1}{2}at^2 \), we have:
\( x = 0 + 0 * t + \frac{1}{2} a t^2 \)
Simplifying, we get: \( x = \frac{1}{2} a t^2 \)
b) For the later phase:
Since the acceleration changes sign, we can use the same equation \( x = x_0 + v_0t + \frac{1}{2}at^2 \), but with \( a \) replaced by \( -a \):
\( x = 0 + 0 * t + \frac{1}{2} (-a) t^2 \)
Simplifying, we get: \( x = -\frac{1}{2} a t^2 \)
4. Equating the positions:
To find the time \( t_1 \) when the point mass returns to its initial position, we set the positions equal to each other:
\( \frac{1}{2} a t_1^2 = -\frac{1}{2} a t_1^2 \)
5. Solving for \( t_1 \):
Simplifying the equation, we get:
\( a t_1^2 = -a t_1^2 \)
Dividing both sides by \( a \), we have:
\( t_1^2 = -t_1^2 \)
Since \( t_1^2 \) cannot be negative, we find that the only solution is when \( t_1^2 = 0 \).
6. Conclusion:
Since \( t_1^2 = 0 \), we can conclude that \( t_1 = 0 \). Therefore, the point mass returns to its initial position at \( t = 0 \) (the instant of starting the motion).
LOL !!!
This is a trick question.
You know the answer.
It is exactly the same as throwing a ball straight up.
It takes just as long to go up
(0 = Vi - g t) so t = Vi/g
h = Vi t - (1/2) g t^2
as it takes to go down
0 = h - (1/2) g t^2
= Vi t - (1/2) g t^2 - (1/2) g t^2
t = Vi / g again going down
so t1 = 2 t