Please solve the problem and show brief steps .

(1 point) The position at time t of a particle that moves along a straight line is given by the function 30'). The first derivative of s(t) is called the velocity, denoted by 00'); that is, the velocity is the rate of change of the position. The rate of change of the velocity is called acceleration, denoted by a(r); that is, Eva) = (1(1). Given that v(r) = 5"(1'), it follows that d2 $50) = 0(1). Find the velocity and acceleration at time I = 1 for the following position functions. 3(1) = 12 8: Velocity at t = 1 is Acceleration att = 1 is 5(1) = V12 + 2 Velocity at r = 1 is Acceleration att = 1 is s(t) = r9 3: Velocity att = 1 is Acceleration at: = 1 i5