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

B)

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

C)


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