Joselyn took her little sister Tina to a garden party on a sunny day. Joselyn wa

Question

Joselyn took her little sister Tina to a garden party on a sunny day. Joselyn was standing 22m from the edge of the pool, while Tina was playing 14m from Joselyn on a straight line between Joselyn and the pool. All the sudden, Joselyn noticed that Tina started running at constant spend of 1.5m/s towards the pool. At that moment, joselyn who was standing started running at constant acceleration. I) How long will it take Tina to reach the edge of the pool? 2) If Joselyn catches Tina just at the edge of the pool, what would her acceleration be? 3) What is Joselyn's speed when she catches Tina at the edge of the pool? 4) Since Joselyn took Physics 1 and got an A, she didn't want to take any chances and started running with an acceleration of 2m/s^2. How far from the edge of the pool will Joselyn catch Tina?

Explanation / Answer

Part 1)

Apply d = vt

14 = 1.5t

t = 9.33 sec

Part B)

d = vot + .5at2

22 = (0) + .5(a)(9.33)2

a = .505 m/s2

Part C)

vf = vo + at

vf = 0 + (.505)(9.33)

vf = 4.72 m/s

Part D)

For Jocelyn d = vot + .5at2

dj = 0 + .5(2)(t2)

For Tina

dt = vt = 1.5t

Also Jocelyn will have a distance of Tina plus 8 m

Thus dt + 8 = .5(2)(t2)

or 1.5t + 8 = t2

In standard form...

t2 - 1.5t - 8 = 0

Find the roots for the time in the quadratic equation

The roots are 3.16 sec and -2.17 sec The negative root can be eliminated)

Thus the time for each will be 3.16 sec

In 3.16 sec, Tina will run (1.5)(3.16) = 4.74 m

Therefore she is 14 - 4.74 = 9.26 m from the pool edge.

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