A car moves along a straight road in such a way that its velocity (in feet per second) at any time t (in seconds) is given by
v(t) = 3t * sqrt(49-t^2)
and ... ?
To find the acceleration of the car at a specific time, we need to take the derivative of its velocity function, v(t). The derivative of v(t) with respect to t will give us the acceleration function, a(t).
To find the derivative of v(t) = 3t * sqrt(49 - t^2), we can use the product rule for differentiation.
The product rule states that if we have two functions u(t) and v(t), then the derivative of their product is given by:
(d/dt)(u(t) * v(t)) = u'(t) * v(t) + u(t) * v'(t)
In this case, u(t) = 3t and v(t) = sqrt(49 - t^2). Let's find the derivatives of both functions:
u'(t) = 3 (since the derivative of t^1 is 1)
To find the derivative of v(t) = sqrt(49 - t^2), we can use the chain rule, since we have a composite function (sqrt inside sqrt).
Let's define g(t) = 49 - t^2. The derivative of g(t) with respect to t is g'(t) = -2t. Now, applying the chain rule:
v'(t) = (1 / (2 * sqrt(g(t)))) * g'(t)
= (1 / (2 * sqrt(49 - t^2))) * (-2t)
= -t / sqrt(49 - t^2)
Now, let's substitute u'(t) = 3 and v'(t) = -t / sqrt(49 - t^2) into the product rule formula:
a(t) = u'(t) * v(t) + u(t) * v'(t)
= 3 * sqrt(49 - t^2) + 3t * (-t / sqrt(49 - t^2))
= 3 * sqrt(49 - t^2) - 3t^2 / sqrt(49 - t^2)
Therefore, the acceleration function of the car at any time t is a(t) = 3 * sqrt(49 - t^2) - 3t^2 / sqrt(49 - t^2).