It is known that from t=2 to t=10 the acceleration of aparticle is inversely pro

Question

It is known that from t=2 to t=10 the acceleration of aparticle is inversely proportional to the cube of the time t. Whent=2 s, v= -15 m/s, and when t =10 s, v=0.36 m/s. Knowing that theparticle is twice as far from the origin when t=2 s as it is when t=10 s, determine: a) the position of the particle when t=2 s and when t=10s, b)the total distance traveled by the particle from t=2 s tot=10 s. It is known that from t=2 to t=10 the acceleration of aparticle is inversely proportional to the cube of the time t. Whent=2 s, v= -15 m/s, and when t =10 s, v=0.36 m/s. Knowing that theparticle is twice as far from the origin when t=2 s as it is when t=10 s, determine: a) the position of the particle when t=2 s and when t=10s, b)the total distance traveled by the particle from t=2 s tot=10 s.

Explanation / Answer

   given   acceleration   a   =   k* t3         where   k   isproportionality constant    velocity   v   =   a * dt   =   (k *t4/4)   +   A            A   =   constantof integration    position   x   =   v * dt   =   (k * t5/ 20)   +   A *t   +   B         B   =   Anotherconstant of integration    at   t   =   2,   v2   =   -15 m/s       -15   =   (k * 24 /4)   +   A   =>   A+ 4 *k   =   -15         ---------(1)    at   t   =   10,   v10   =   0.36 m/s    0.36   =   (k* 104 /20)   +   A   =>   A+ 500 *k   =   0.36      --------(2)    Solving equation (1) and(2)      k   =   0.031    and   A   =   -15.124    also   x2    =   2* x10        (k *25 / 20)   +   A* 2   +   B   =    2* {(k * 105 /20)   +   A *10   +   B}    Substituting vaues          1.6 *0.031   +   2 *(-15.124)   +   B   =   10000* 0.031 + (-15.124) *20   +   2 * B    B   =   -37.72    a.   x2   =   (k *25 / 20)   +   A* 2   +   B   =   0.031* 1.6   +   (-15.124 *2)   -   37.72   =   -67.92   m          x10   =   {(k *105 / 20)   +   A *10   +   B}   =   0.031* 5000   +   ( -15.124 * 10) -37.72   =   -33.97   m    b.   distancetravelled   x   =   x10   -   x2   =   -33.97 - ( -67.92)   =   33.95   m        (k *25 / 20)   +   A* 2   +   B   =    2* {(k * 105 /20)   +   A *10   +   B}    Substituting vaues          1.6 *0.031   +   2 *(-15.124)   +   B   =   10000* 0.031 + (-15.124) *20   +   2 * B    B   =   -37.72    a.   x2   =   (k *25 / 20)   +   A* 2   +   B   =   0.031* 1.6   +   (-15.124 *2)   -   37.72   =   -67.92   m          x10   =   {(k *105 / 20)   +   A *10   +   B}   =   0.031* 5000   +   ( -15.124 * 10) -37.72   =   -33.97   m    b.   distancetravelled   x   =   x10   -   x2   =   -33.97 - ( -67.92)   =   33.95   m

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