Question

Bob is standing at rest. Wendy runs past him at a constant speed of 3.00 m/s. At the instant that Wendy passes him, Bob starts to run after her with a constant acceleration of 0.200 m/s How far must Bob run before he catches up with Wendy

Answer 1

Answer:

90 meters

Explanation:

To find the distance in which Bob catches up with Wendy, you first write down the motion equations of both Bob and Wendy.

Wendy has a constant speed, then you have:

$x_1=v_1t$           (1)

Bob has an accelerated motion, then, his equation of motion is:

$x_2=v_2t+\frac{1}{2}at^2$          (2)

v1: constant speed of Wendy = 3.00 m/s

v2: initial speed of Bob = 0 m/s

a: acceleration of Bob = 0.200m/s^2

When Bob catches up with Wendy x1 = x2, then you equal both equations and solve for time t:

$x_1=x_2\\\\v_1t=\frac{1}{2}at^2\\\\t=\frac{2v_1}{a}$

you replace the values of a and v1:

$t=\frac{2(3.00m/s)}{0.200m/s^2}=30 s$

Finally, you replace this value of time either the equation for x1 or the equation for x2, and you calculate the distance:

$x_2=x_1=(3.00m/s)(30s)=90m$

hence, Bob catches up with Wendy for a distance of 90m