Suppose that the equation of motion for a particle (where s is in meters and t in seconds) is s=(1/3)t3−7t2+49t+4 . (a) Find the velocity and acceleration as functions of t. Velocity at time t = t^2-14t+49 Acceleration at time t = dne (b) Find the acceleration after 1 second. Acceleration after 1 second: dne (c) Find the acceleration at the instant when the velocity is 0. Acceleration: 0

I'll do it rn and then answer

Since you have not included the experiment situation and the data, of course, i can not give you specific details about it. but i can you with this question.1) i will assume that the motion is projectile (parabolic) motion.2) yes, the horizontal distance depends on the horizontal velocity as, naming vx the horizontal velocity and t the time, it is: x = vx × t3) now, observe that the flight time depends on the vertical velocity as, naming voy the initial vertical velocity, t the time and y the vertical distance before reaching the ground: y = yo + voy×t - gt² / 2.the higher the initial vertical velocity (voy) the longer the object will be flying and so t will be higher in the equation for the horizontal distance.that explains the relationship between the horizontal distance and the horizontal velocity.