Question

A particle moves on a coordinate line with an acceleration at time t seconds of e^t/2 m/sec^2. At t=0 the particle is at the origin, and its velocity is -4 m/sec

Answer 1

Hello,

A) $a=e^ \frac{t}{2} \\\\ v(t)=\int\ a\ dt=\int\ e^ \frac{t}{2}\ dt=2*e^ \frac{t}{2}+C\\\\v(0)=-4\\ 2*e^ \frac{0}{2}+C=-4\\ C=-6\\\\ \boxed{v(t)= 2*e^ \frac{0}{2}-6}$

B)
$2*e^ \frac{t}{2}-6=0\\\\ e^ \frac{t}{2}=3\\\\ t=2*ln(3)\approx{2,19722...}$

C)

$x(t)=\int_0^6(2*e^ \frac{t}{2}-6})\ dt=\left[4*e^ \frac{t}{2}-6t\right]_0^6\\\\ =e^3-36-4=4e^3-40\approx{40.3421...}$