Answer to Question #206457 in Physics for Morgan Hopkin

Question

Miss Takken is cruising along at 4 m/s (in her cool FizzX car) when a physics student

(who did not hand in their lab) passes her in a pickup truck going at a speed of 13 m/s. 

If Miss Takken can accelerate at 2.5 m/s2

a.     How long will it take Miss Takken to catch this student?

b.     At what distance from the initial pass will she catch them?

c.     How fast will Miss Takken be going when she catches the student?

Answer 1

(a) Let's first write the distance traveled by the student:

$$d_1=v_{01}t.$$

Let's write the distance traveled by Miss Takken:

$$d_2=v_{02}t+\dfrac{1}{2}at^2.$$

Miss Takken will catch this student when $d_1=d_2$:

$$v_{01}t=v_{02}t+\dfrac{1}{2}at^2,$$ $$1.25t^2-9t=0,$$ $$t(1.25t-9)=0,$$ $$t=\dfrac{9\ \dfrac{m}{s}}{1.25\ \dfrac{m}{s^2}}=7.2\ s.$$

Miss Takken will catch this student after $t=7.2\ s$.

(b) We can find at what distance from the initial pass will she catch the student from the kinematic equation:

$$d_2=v_{02}t+\dfrac{1}{2}at^2,$$ $$d_2=4\ \dfrac{m}{s}\cdot7.2\ s+\dfrac{1}{2}\cdot2.5\ \dfrac{m}{s^2}\cdot(7.2\ s)^2=93.6\ m.$$

(c) We can find how fast will Miss Takken be going when she catches the student from another kinematic equation:

$$v=v_0+at,$$ $$v=4\ \dfrac{m}{s}+2.5\ \dfrac{m}{s^2}\cdot7.2\ s=22\ \dfrac{m}{s}.$$