Displacement i sgivenby x=2+3t+4t2. Find the value of instantaneous acceleration.
x'=3+8t
x"=a=8
Sorry, I typed the equation wrong. This is the correct one.
"Displacement is given by x=2+3t+4t². Find the value of instantaneous acceleration."
I still don't understand how to go about this.
Thanks.
To find the instantaneous acceleration, we need to take the derivative of the displacement equation with respect to time (t).
Given that the displacement equation is x = 2 + 3t + 4t^2, let's proceed to find the derivative:
First, let's differentiate each term of the equation with respect to t:
d(x)/dt = d(2)/dt + d(3t)/dt + d(4t^2)/dt
The derivative of a constant (2) with respect to t is zero, since a constant does not change with respect to time. Similarly, the derivative of 3t with respect to t is 3, since t is raised to the power of 1. Then, the derivative of 4t^2 with respect to t is 8t, following the power rule of differentiation.
Simplifying the terms, we have:
d(x)/dt = 0 + 3 + 8t
Therefore, the derivative of the displacement equation is:
dx/dt = 3 + 8t
This expression represents the instantaneous velocity of the object. To find the instantaneous acceleration, we need to take the derivative of the velocity equation with respect to t:
d(dx/dt)/dt = d(3 + 8t)/dt
Again, we differentiate each term with respect to t:
d(dx/dt)/dt = d(3)/dt + d(8t)/dt
The derivative of a constant (3) with respect to t is zero, and the derivative of 8t with respect to t is 8, since t is raised to the power of 1.
Simplifying the terms, we have:
d(dx/dt)/dt = 0 + 8
Therefore, the derivative of the velocity equation, which is the acceleration equation, is:
d^2x/dt^2 = 8
Hence, the value of instantaneous acceleration is 8.