If a particle's motion is described as x = ut+1/2at^2, where x is position, t is time and u and a, are constants. Show that the acceleration of the particle is constant
To show that the acceleration of the particle is constant, we need to differentiate the equation for position with respect to time.
Given: x = ut + (1/2)at^2
Differentiating both sides of the equation with respect to time (t), we get:
d/dt(x) = d/dt(ut + (1/2)at^2)
On the left side, we differentiate x with respect to t, which gives us the velocity (v). On the right side, we differentiate each term separately.
v = u + at (Differentiating ut with respect to t gives us u, and differentiating (1/2)at^2 with respect to t gives us at).
Now, let's differentiate the equation for velocity (v) with respect to time (t) to find the acceleration (a).
d/dt(v) = d/dt(u + at)
On the left side, we differentiate v with respect to t, which gives us acceleration (a). On the right side, as u and a are constants, their derivatives with respect to t are zero.
a = 0 + a
Therefore, the acceleration (a) of the particle is constant.
In summary, by differentiating the equation for position (x) with respect to time, we obtain the equation for velocity (v), and by differentiating the equation for velocity with respect to time, we obtain the equation for acceleration (a). In this case, since the equation for acceleration does not include any terms with t (time), the acceleration remains constant.