Question

1.     A body moving with uniform acceleration has velocity of 1.2m/s when its coordinate is 300cm. The x coordinate of a body 2s later is 600cm.

(a)  What is the magnitude of its acceleration?

(b)  Find the instantaneous velocity at time t = 3s.

Answer 1

(a) We can find the magnitude of its acceleration from the kinematic equation:

$$x_f-x_i=v_0t+\dfrac{1}{2}at^2,$$ $$a=\dfrac{2(x_f-x_i)-2v_0t}{t^2},$$ $$a=\dfrac{2\cdot(6\ m-3\ m)-2\cdot1.2\ \dfrac{m}{s}\cdot2\ s}{(2\ s)^2}=0.3\ \dfrac{m}{s^2}.$$

(b) The instantaneous velocity of the body is the derivative of $x$ with respect to $t$:

$$v(t)=\dfrac{d}{dx}(x(t)),$$ $$v(t)=\dfrac{d}{dx}(x_0+v_0t+\dfrac{1}{2}at^2)=v_0+at,$$ $$v(t=3\ s)=1.2\ \dfrac{m}{s}+0.3\ \dfrac{m}{s^2}\cdot3\ s=2.1\ \dfrac{m}{s}.$$