You have the wrong integral. Distance is the integral of velocity with respect to time, not with respect to distance.rudransh verma said:Homework Statement:: A particle at x=0 at time t=0 starts moving along positive x-axis with velocity v= alpha## \sqrt x##. Displacement of particle is
Relevant Equations:: ##v=\frac{dx}{dt}##.
Integration of v= integration of##(alpha \sqrt x)dx##.
But I am getting wrong answer.
The concept of integration of velocity to get displacement is based on the fundamental relationship between velocity and displacement. Velocity is the rate of change of displacement over time, and integration is the reverse process of differentiation. By integrating the velocity function over a specific time interval, we can obtain the displacement traveled during that time period.
Integrating velocity to get displacement is necessary because it allows us to determine the total distance traveled by an object over a specific time period. This is particularly useful when dealing with non-constant velocities, as the displacement cannot be calculated simply by multiplying the velocity by time.
The integration of velocity to get displacement is directly related to the area under a velocity-time graph. This is because the integral of velocity is equal to the displacement, and the area under the curve of a velocity-time graph represents the displacement traveled during a specific time interval.
The units for displacement obtained through integration of velocity are typically the same as the units for velocity, but with an additional unit of time. For example, if the velocity is measured in meters per second, the displacement will be measured in meters.
Yes, integration of velocity can be used to calculate displacement for any type of motion, as long as the velocity function is known. This includes both linear and non-linear motion, as well as motion with changing velocities. However, it is important to note that the accuracy of the calculated displacement may depend on the accuracy of the measured velocity data.