Question

A partial move in straight line according to law v= 2 (xsinx + coax) .find its acceleration at X=π/2

Answer 1

Here is the answer .

Answer 2

Answer:

a = 1units²

Explanation:

In this case, we are given velocity as a function of 'x', which we can say as time(not sure as it is not mentioned).

Now, for acceleration, we know that acceleration is nothing but the rate of change of velocity.

In simple words, to determine acceleration, take the derivative of velocity at that scale;

Thus;

a = dv/dx

Where,

'x' shows time in this case, not the displacement.

Now,to answer about this 'x', it must be time as if we differentiate 'v' with respect to displacement, we cannot achieve acceleration as it is the derivative of velocity with respect to time.

Thus; differentiating as;

a = dv/dx = d/dx(xsinx+cosx)

Now by theorem of derivatives;

d/dx(f(x)+g(x)) = d(f(x))/dx + d(g(x))/dx

Where,

f(x) and g(x) are two different functions.

Now, we know;

d/dx(f(x)×g(x)) = f(x)×d(g(x)/dx + g(x)×d(f(x))/dx (product rule for two function for derivative).

And;

d(sinx)/dx = cosx, and;

d(cosx)/dx = -sinx.

a = (xcosx+sinx) + (-cosx)

a = xcosx + sinx -cosx

Thus, for x = pi/2;

a(at x = pi/2) = (pi/2)cos(pi/2) + sin(pi/2) - cos(pi/2)

Thus; putting cos(pi/2) = 0 and sin(pi/2) = 1

a = (pi/2)(0)+1-0

a = 1 units²