if a = 4 - t^2 , then
v = 4t - (1/3)t^3 + c
given: when t=0, v = 2 mm/s
2 = 0 - 0 + c
c = 2
v = 4t - (1/3)t^2 + 2
s = 2t^2 - (1/12)t^4 + 2t + k
when t=0 , s = 0 , so k = 0
s = 2t^2 - (1/12)t^4 + 2t
check by taking the derivative twice
Integrate two times and substitute the initial conditions to find the required equation:
From x'(t)=2 mm-t^-1, we get
2 mm-t^-1 = 4(0)-(0^3/3)+C1
x(0)=0 => C2=0
x(t)=2t² -(1/12)t^4 +2t
dynamics - The acceleration of a particle is defined by the relation a = -60x^-1...
math(mechanics) - a particle moves along a straight line such that its position ...
physics,dynamics - The acceleration of a particle along a straight line is ...
math - A particle is traveling along a one-dimensional path (such as a number ...
Physics (Algerbra-based) - A particle is moving clockwise around a circle with a...
AP Calculous - A particle moves on the x ľaxis so that its position at any time...
calculous - A particle moves on the x ľaxis so that its position at any time is ...
HS Calculus - Having trouble with this questions. Please help. A particle moves ...
physics - Two particles A and B are in uniform circular motion about a common ...
Calculus - The position of a particle moving on a horizontal line is given by s(...