The acceleration of a particle is defined by the following relation where A is a

Question

The acceleration of a particle is defined by the following relation where A is a constant At t = 0 the position of the particle is 8 meters measured from the origin and is at rest. At the end of the first second the velocity of the particle is 30 meters per second. Note: Distances are in meters and time in seconds. Hint; Find the expression of velocity and acceleration as a function of time. What is the value of the constant A? What is the position of the particle at the end of 2 seconds? What is the velocity of the particle at the end of 3 seconds? What is the acceleration of the particle at the end of 3 seconds?

Explanation / Answer

a = A - 6 t^2
dv/dt = A - 6 t^2
dv =( A - 6 t^2) dt
integrating both sides
v= At - 6t^3 / 3 + C
since at t=0 , v=0 so
0 = A(0) - 6(0)^3 + C
or C = 0
v(t) = At - 6t^3 / 3
also at t = 2 s , v =30
so 30 = A(2) -(6/3) * 2^3
or 2A - ( 8/3) 6 = 30 ;
0r A = 23 ;
or dx /dt = (At - 6t^3 / 3 )
or dx = (At - 6t^3 / 3 ) dt
integrating both sides
x = At^2 /2 - 6t^4 / 12 + C1
since at t=0 , x = 8 so
8 = 0 - 0 + C1
or C1 = 8 ;
so x( t ) = At^2 /2 - 6t^4 / 12 + 8

 

(a ) value of A = 23 ;

 

(b) at the end of 2 sec , t= 3 s
so x(t) = x( 3 ) = 23 * 9 /2 - 3^4 / 2 + 8;
x = 71 m

 

(c) at the end of 3 s , t= 4
v ( t) = At - 6t^3 / 3 = 23 * t - 6t^3 / 3
v( 4 ) = -36 m /s

 

(d) a = A - 6 t^2 = a = 23 - 6 t^2 ;
a(4) = -73 m /s^2

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