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

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Answer 1 · 2016-05-27T20:02:50

$the position of the particle at t=5 is \frac{25}{6} m$ $"the position of the particle at t=5 is "25/6" m"$

$using diagram$ $"using diagram"$
$a (t) = a_{i} - k \cdot t$ $a(t)=a_i-k*t$

$a_{i} = 6$ $a_i=6$
$k = tan α = \frac{6}{6} = 1$ $k=tan alpha=6/6=1$

$a (t) = 6 - t$ $a(t)=6-t$

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

$v (t) = 6 t - \frac{1}{2} t^{2} + C$ $v(t)=6t-1/2t^2+C$

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

$v (t) = 6 t - \frac{1}{2} t^{2} + 15$ $v(t)=6t-1/2t^2+15$

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

$x (t) = \frac{1}{2} \cdot 6 \cdot t^{2} - \frac{1}{2} \cdot \frac{1}{3} \cdot t^{3} + 15 t + C$ $x(t)=1/2*6* t^2-1/2*1/3*t^3+15t+C$

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

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

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

$x (5) = 3 \cdot 5^{2} - \frac{1}{6} \cdot 5^{3} + 15 \cdot 5 + 25$ $x(5)=3*5^2-1/6* 5^3+15*5+25$

$x (5) = 75 - \frac{125}{6} + 75 + 25$ $x(5)=75-125/6+75+25$

$x (5) = 25 - \frac{125}{6}$ $x(5)=25-125/6$

$x (5) = \frac{150 - 125}{6}$ $x(5)=(150-125)/6$

$x (5) = \frac{25}{6}$ $x(5)=25/6$