this is one of many sample problems i have for homework and i dont understand them at all, can someone tell me how to do it so i can atleast try the rest? We have a test on this friday and i would like to know what im doing. thankss

- A speeder driving down the road at a constant 20 m/s, passes a patrolman parked roadside. The patrolman waits 3 secs, then pursues the speeder accelerating at a constant 4.0m/s(squared).

a. How long does it take for the patrolman to catch the speeder?

b. How far has he traveled before doing so?

## Answers (2)

Distance the speeder is from the patrol when the pursuit begins (i.e. the speeder's head start in terms of distance)

20m/s * 3s = 60m

Distant traveled by the speeder between being spotted and being caught

d = v_s * t + 60m

Distant traveled by the patrol between spotting the speeder and catching him

d = 1/2*a*t^2 + vi*t + d_0 = 1/2*a*t^2 (since the other 2 terms are both 0)

Equate the 2 expressions for d:

v_s*t + 60m = 1/2*a*t^2

1/2*a*t^2 - v_s*t - 60m = 0

1/2*(4m/s^2)*t^2 - (20m/s)*t - 60m = 0

Now use the quadratic equation to solve for t, with a = 2 , b = -20, c = -60

The positive answer (i.e. the only answer that makes sense in a time sense) is t = 12.42s

So that is the answer to part a.

for part b, the value for t can be plugged back into either equation for the d:

d = 308.3m

speeder is initially 3 sec ahead - at 20 m/s this is a distance = (3)(20) = 60 m

in order to gain 60 m on the constant velocity speeder, the patrolman in the

SAME time = t must travel:

60 + 20t = 1/2at² {where a = 4.0}

60 + 20t = (0.5)(4.0)t²

0 = 2.0t² - 20t - 60

solve for positive root of t:

t = 12.4 s ANS

substitute t into either (60 + 20t) or (1/2at²) to find ans for b.