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: Calculate time for dune - buggy to reach the car

The dune - buggy moves at a constant speed $v_{buggy}=0.57$ m/s and travels a distance $d = 2.55 + 3.65=6.2$ m. Using the formula $d = v\times t$, we can find the time $t$.
$t=\frac{d}{v_{buggy}}$
$t=\frac{6.2}{0.57}\approx 10.88$ s

Step2: Calculate 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. Using the kinematic equation $d_{car}=v_{0}t+\frac{1}{2}at^{2}$, we can solve for $a$.
$3.65 = 1.05\times10.88+\frac{1}{2}a\times(10.88)^{2}$
$3.65=11.424 + 59.19a$
$59.19a=3.65 - 11.424$
$59.19a=- 7.774$
$a=\frac{-7.774}{59.19}\approx - 0.13$ m/s²

Answer:

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