Question

The diaplacement of a particle in X direction vary with time given by X=5t square+6t-2. Find the initial velocity of the particle?

Answer 1

Answer:

Velocity =6m/s

Explanation:

We have x=5t^2 + 6t -2. Differentiate it with respect to time so we get dx\dt = 10t + 6.

we know that when we differentiate distance with respect to time we get velocity, therefore V= 10t+6. Since initial velocity is asked time becomes zero.

so, V= 10×0+6

Therefore, V = 6m/s

Answer 2

Answer:

Dx/Dt = 10t+6

Dx/Dt becomes V

V= 10×0+6

V= 6m/s

Explanation:

Step 1: You gotta differentiate it. Divide x(distance) with t(time) so that it becomes v(velocity).

Step 2: the power on t come down and gets multiplied with 5 and the power on t itself minuses with 1 thus it becomes 1.

Step 3: the power on 6t is 1 so minuses with 1 thus becomes 0, so only 6 remains.

Step 4: -2 according to the rules of differentiation variables and numbers (without any variables joined with it) that are alone becomes zero

Step 5: initial time velocity is 0 sec, so substituting 0 in t gives you 6m/s. Voila!