SOLUTION: A particle travelling in a straight line passes a fixed point O with a velocity of 16 m/s. Its acceleration, a m/s^2, is given by a = 12- 6t, where t is the time in seconds after

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Question 1195353: A particle travelling in a straight line passes a fixed point O with a velocity of 16 m/s. Its
acceleration, a m/s^2, is given by a = 12- 6t, where t is the time in seconds after passing O. Calculate
a) the greatest speed attained by the particle in the original direction of the motion.
b) The distance from O when t = 2

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
A particle travelling in a straight line passes a fixed point O with a velocity of 16 m/s. Its
acceleration, a m/s^2, is given by a = 12- 6t, where t is the time in seconds after passing O. Calculate
a) the greatest speed attained by the particle in the original direction of the motion.
b) The distance from O when t = 2
a%22%22=%22%2212-6t

dv%2Fdt%22%22=%22%2212-6t

dv%22%22=%22%2212%2Adt%22%22-%22%226t%2Adt%29

int%28dv%29%22%22=%22%2212%2Aint%28dt%29%22%22-%22%226%2Aint%28t%2Adt%29

v%22%22=%22%2212t%22%22-%22%223t%5E2%22%2B%22constant

When t=0, the particle is at O, with velocity 16. So we substitute
t=0 and v=16

16%22%22=%22%2212%280%29%22%22-%22%223%280%29%5E2%22%2B%22constant

So the constant is 16, and

v%22%22=%22%2212t%22%22-%22%223t%5E2%22%2B%2216

Use the vertex formula or differentiating and setting the derivative = 0
to find that the maximum velocity is 

v=28 m/s when t=2 seconds.        <--answer to (a)   

To find the distance when t=2,

v%22%22=%22%2212t%22%22-%22%223t%5E2%22%2B%2216

ds%2Fdt%22%22=%22%2212t%22%22-%22%223t%5E2%22%2B%2216

ds%22%22=%22%2212t%2Adt%22%22-%22%223t%5E2%2Adt%22%2B%2216%2Adt

int%28ds%29%22%22=%22%2212int%28t%2Adt%29%22%22-%22%223int%28t%5E2%2Adt%29%22%2B%2216int%28dt%29


s%22%22=%22%226t%5E2%22%22-%22%22t%5E3%29%22%2B%2216t%22%22%2B%22%22constant

When t=0, the particle is at O, so at that instant the distance s=0, So
we substitute t=0 and s=0

0%22%22=%22%226%280%29%5E2%22%22-%22%22%280%29%5E3%29%22%2B%2216%280%29%22%22%2B%22%22constant

So the constant is 0

s%22%22=%22%226t%5E2%22%22-%22%22t%5E3%29%22%2B%2216t

We substitute t=2

s%22%22=%22%226%282%29%5E2%22%22-%22%22%282%29%5E3%29%22%2B%2216%282%29

s = 48 meters.       <--answer to (b)

Edwin