At $t=0$ , a particle is located at $x=25m$ and has a velocity of $15m/s$ in the positive $x$ direction. The acceleration of the particle varies with time as shown in the diagram. What is the position of the particle at $t=5.0s$ ?

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Answer 1 · 2016-05-27T20:02:50

$"the position of the particle at t=5 is "25/6" m"$

$"using diagram"$
$a(t)=a_i-k*t$

$a_i=6$
$k=tan alpha=6/6=1$

$a(t)=6-t$

$v(t)=int (6-t)d t$

$v(t)=6t-1/2t^2+C$

$"for t=0 v=15 m/s C=15"$

$v(t)=6t-1/2t^2+15$

$x(t)=int v(t) d t=int (6t-1/2 t^2+15) d t$

$x(t)=1/2*6* t^2-1/2*1/3*t^3+15t+C$

$x(t)=3t^2-1/6t^3+15t+C$

$"for t=0 x=25 m, C=25$

$x(t)=3t^2-1/6x^3+15t+25$

$x(5)=3*5^2-1/6* 5^3+15*5+25$

$x(5)=75-125/6+75+25$

$x(5)=25-125/6$

$x(5)=(150-125)/6$

$x(5)=25/6$