a point mass starts moving in a straight line with constant acceleration "a". at a point after the beginning of motion,the acceleration changes sign, without change in magnitude. determine time T from the beginning of the motion in which the point mass returns to the initial point.
Answers
s = u t + 1/2 a t² ,
u = 0 and at time t1, when the acceleration changes, distance travelled:
s = 1/2 a t1²
v at t = t1 = u + a t = a t1
Now the acceleration is changed to -a. Then the particle continues in the same direction until the velocity becomes zero. Then the particle changes the direction and starts accelerating and passes over the point of start.
u = a t1 v = 0 acceleration = -a
v = u + a t
=> 0 = a t1 - a t => t = t1 it takes t1 more time to stop and reverse direction.
The distance traveled/displacement in this time:
s = u t + 1/2 a t²
=> s = a t1 * t1 - 1/2 a t1² = 1/2 a t1²
The total displacement from the initial point : 1/2 a t1² + 1/2 a t1² = a t1²
now, acceleration = -a u = 0 s = - a t1² in the negative direction
s = u t + 1/2 a t²
=> - a t1² = 0 - 1/2 a t²
=> t = √2 t1
The total time T from initial point forward till back to initial point :
T = 2 t1 + √2 t1 = (2 + √2) t1
Answer:
Explanation:
Using the second equation of motion- s = ut + 1/2 at²
where u = 0 and time = t1
When the acceleration changes, the distance travelled will be:
= s = 1/2 a t1²
When the acceleration changes to -a, then the particle continues in the same direction until the velocity will become zero. There after the particle changes the direction, starts to accelerate and passes over the point of start.
Using first equation of motion - v = u + a t
where v = 0 and acceleration = -a
= 0 = a t1 - a t
= t = t1 sincw, it takes t1 more time to stop and reverse direction.
Hence, the distance traveled/displacement in this time:
= s = u t + 1/2 at²
= s = a t1 ×t1 - 1/2 at1² = 1/2 a t1²
Total displacement from the initial point : 1/2 a t1² + 1/2 a t1² = a t1²
In negative direction - s = u t + 1/2 a t²
= -a t1² = 0 - 1/2 a t²
= t = √2 t1
The total time T from initial point forward till back to initial point is
T = 2 t1 + √2 t1
= (2 + √2) t1.