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I have some problems >?
1) The velocity of a particle is given by v=100 + 280e^-kt ms^-1 . If the velocity decreases by 20% after 50 seconds , find the % decrease in velocity after 3 minutes .
2)The formula for the acceleration of a particle is given by A=5x-e^2x +3 where x is the displacement of the particle . If the acceleration of the particle is at a constant rate of -9.8ms^-2 , find the rate of change of its displacement when the displacement is -0.3m .
please show me ur working out
2 Answers
- ted sLv 79 years agoFavorite Answer
1. ≈ 73.4% decrease...v(50 ) = 0.8 v(0) = 8 * 38 ≡ 100 + 280 e^(-50k)---->
ln [ ( 8*38 -100) / 280 ] = - k50----> k ≈ 15.83---> v(300) ≈ 100--->(280/380) ≈ .734
#2...makes little sense...you 1st imply that acceleration is dependent upon position
and then 2nd you say it is constant !!...which ??
I will , for the sake of doing some work , assume that at some value for x
that the acceleration is - 9.8 and we wish to know the velocity at that point...
these x values are ≈ 1.55 and - 2.55.....v = 2.5 x² - 0.5 e^(2x) + 3x + C...v(0) = -0.5 + C
thus v(x) = 2.5 x² - 0.5 e^(2x) + 3x + 0.5 + v(0)...now let x = one of the values
- Mark S, JPAALv 79 years ago
This is an educated guess for the first one:
Take the derivative of the equation to obtain dV/dt. Then set this equal to 0.2 (which is 20%), substitute 5/6 (50 sec, which is 5/6 of a minute) for t and solve for k. When you have that, substitute 3 in for t and solve to get dV/dt after 3 min.
