Answer to Question #164720 in Physics for jaemin

Question

accelerates uniformly from a velocity of 1 m s^-1 to a velocity of 5 m s^-1 in 0.5 minutes . Calculate the displacement of the transporter .

Answer 1

By definition, the average acceleration is given as:

$$a = \dfrac{v_2-v_1}{t}$$

where $v_1,v_2$ are initial and final velocities respectively, and $t$ is the time interval the object experienced the acceleration. Substituting our values, obtain:

$$a = \dfrac{5m/s-1m/s}{0.5s} = 8m/s^2$$

The distance traveled under the constant acceleration is given as:

$$d = v_1t + \dfrac{at^2}{2}\\ d = 1m/s\cdot 0.5s + \dfrac{8m/s^2\cdot (0.5s) ^2}{2} = 1.5m$$

Answer. 1.5 m.