1. A speedboat moving at 30.0 m/s approaches a no-wake buoy marker 100 m ahead.

Question

1. A speedboat moving at 30.0 m/s approaches a no-wake buoy marker 100 m ahead. The pilot slows the boat with a constant acceleration of -3.3 m/s2 by reducing the throttle.

(a) How long does it take the boat to reach the buoy (in seconds)?

(b) What is the velocity of the boat when it reaches the buoy (in m/s)?

2. The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of -5.35 m/s2 for 4.10 s, making straight skid marks 60.9 m long, all the way to the tree. With what speed does the car then strike the tree (in m/s)?

Explanation / Answer

First working formula,

S = VT - (1/2)aT^2

where

S = distance of boat from buoy = 100 m

V = initial velocity of boat = 30 m/sec

T = time for boat to reach the buoy

a = acceleration = - 3.3 m/sec^2

Substituting values,

100 = 30T - (1/2)(3.3)T^2

100 = 30T - 1.65T^2

Rewriting the above,

1.65T^2 - 30T + 100 = 0

Using the quadratic formula,

T = 4.396 sec. ...................ans

Next working formula is

Vf - Vo = aT

where

Vf = velocity of boat as it reaches the buoy

and all the other terms have been defined.

Substituting values,

Vf - 30 = (-3.3)(4.396)

Vf = 15.49 m/sec. .....ans

2)dist = v0t+1/2at^2 to find initial speed

vf=final speed (to be found)

v0=initial speed (need to find)

a=accel = -5.35m/s/s

d=dist = 60.9

first equation gives us:

60.9=v0(4.1)-1/2(5.35)(4.1)^2

v0=25.82m/s

if the initial speed is 25.82 m/s, and the car loses speed at the rate of 5.35m/s/s for 4.1s, then the car loses 5.35x4.1 = 21.935m/s and thus hits the tree with a speed of

25.82 - 21.935=3.885m/s

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