Answer to Question #244643 in Physics for Lol

Question

A jogger accelerates from rest to 5.13 m/s in 3.38 s. A car accelerates from 23.0 to 35.2 m/s also in 3.38 s. (a) Find the magnitude of the acceleration of the jogger. (b) Determine the magnitude of the acceleration of the car. (c) How much further does the car travel than the jogger during the 3.38 s?

Answer 1

(a) The acceleration is

$$a_j=\frac{v_f-v_i}{t}=1.52\text{ m/s}^2.$$

(b) The acceleration is

$$a_c=\frac{v_f-v_i}{t}=3.61\text{ m/s}^2.$$

(c) The difference in their displacements is

$$\Delta d=d_c-d_j=\bigg(v_it+\frac{a_ct^2}{2}\bigg)-\bigg(\frac{a_jt^2}{2}\bigg)=89.7\text{ m}.$$