A particle is put inside an accelerator at time t=0. After t sec, its velocity is 10^5t^2 m/s. How far does the particle move during the first 10**-2 sec?
distance= int velocity dt
= INT 1E5 t^2 dt
limits 0 to 1E-2 sec
To find the distance the particle moves during the first 10^(-2) seconds, we need to integrate the given velocity function with respect to time from 0 to 10^(-2) seconds.
The velocity function is given as 10^5t^2 m/s.
So, the distance can be calculated as follows:
distance = ∫(10^5t^2) dt
To integrate this function, we use the power rule of integration, which states that the integral of t^n is equal to (t^(n+1))/(n+1).
Applying this rule, we get:
distance = (10^5 * (t^2+1)) / (2+1)
Now, we can calculate the distance by substituting the limits of integration:
distance = [(10^5 * ((10^(-2))^2+1)) / (2+1)] - [(10^5 * (0^2+1)) / (2+1)]
Simplifying this equation, we have:
distance = [(10^5 * (10^(-4) + 1)) / 3] - [(10^5 * 1) / 3]
Further simplifying, we get:
distance = [(10^5 * (10^(-4) + 1 - 1)) / 3]
Since 1 - 1 is equal to 0, we can simplify further to:
distance = [(10^5 * (10^(-4))) / 3]
Finally, evaluating this expression, we have:
distance = (10^5 * 10^(-4)) / 3
= (10^1) / 3
= 10/3 meters
Therefore, the particle moves a distance of 10/3 meters during the first 10^(-2) seconds.