the position of a particle moving along x-axis is given=5t^2-10t the distance travelled by particle is
I want an answer
To find the distance traveled by the particle, we need to integrate the absolute value of the velocity function over the given time interval. The velocity function is the derivative of the position function.
First, let's find the velocity function by taking the derivative of the given position function:
v(t) = d/dt (5t^2 - 10t) = 10t - 10
The absolute value of the velocity function will give us the speed of the particle.
Now, let's find the time interval over which we want to calculate the distance. If no time interval is given, we can assume it to be from t=0 to t=tf (final time).
To find the distance traveled, we need to integrate the absolute value of the velocity function over this time interval:
distance = ∫[0 to tf] |v(t)| dt
Let's solve this integral:
distance = ∫[0 to tf] |(10t - 10)| dt
The integral of |(10t - 10)| can be split into cases based on the values of (10t - 10).
Case 1: (10t - 10) ≥ 0, i.e., 10t - 10 ≥ 0, which implies t ≥ 1.
∫[0 to tf] ((10t - 10)) dt = ∫[1 to tf] (10t - 10) dt
= [5t^2 - 10t] from t = 1 to tf
= 5(tf^2 - 1) - 10(tf - 1)
Case 2: (10t - 10) < 0, i.e., 10t - 10 < 0, which implies t < 1.
∫[0 to tf] ((10t - 10)) dt = ∫[0 to 1] (10t - 10) dt
= [-5t^2 - 10t] from t = 0 to 1
= -5(1^2 - 0^2) - 10(1 - 0)
Now, we need to sum the distances from the two cases:
distance = [5(tf^2 - 1) - 10(tf - 1)] + [-5(1^2 - 0^2) - 10(1 - 0)]
Simplifying the expression will give you the distance traveled by the particle.