Question

question 4 (5 points) listen a battery - operated toy dune buggy traveling at a constant speed of 0.57 m/s is 2.55 m behind a wind - up car currently moving at 1.05 m/s but slowing down with a constant acceleration. if the dune buggy catches up with the car 3.65 meters ahead of its current location, what is the acceleration of the car, and how long did it take for the dune buggy to reach it? show all calculations in the answer box.

Explanation:

Step1: Find the time it takes for the dune - buggy to reach the car

The dune - buggy has to travel a distance $d_{buggy}=2.55 + 3.65=6.2$ m. Since it moves at a constant speed $v_{buggy}=0.57$ m/s, we use the formula $d = vt$.
$t=\frac{d_{buggy}}{v_{buggy}}$
$t=\frac{6.2}{0.57}\approx10.88$ s

Step2: Find the acceleration of the car

The car moves a distance $d_{car}=3.65$ m in time $t = 10.88$ s with an initial velocity $v_{0}=1.05$ m/s. We use the kinematic equation $d=v_{0}t+\frac{1}{2}at^{2}$.
$3.65 = 1.05\times10.88+\frac{1}{2}a\times(10.88)^{2}$
$3.65=11.424 + 59.19a$
$59.19a=3.65 - 11.424$
$59.19a=- 7.774$
$a=\frac{-7.774}{59.19}\approx - 0.13$ m/s²

Answer:

The acceleration of the car is approximately $-0.13$ m/s² and the time it takes for the dune - buggy to reach the car is approximately $10.88$ s.