An object's position can be modeled with the following function: s(t) = 2cos(2πt) +t^2, where t is measured in seconds. What will the object's acceleration be after 2 seconds?

A particle moves so that its position (in meters) as a function of time (in seconds) is r = î + 5t2 j + 5t k. (Use the following as necessary: t.) (a) Write an expression for its velocity as a function of time. V = m/s (b) Write an expression for its acceleration as a function of time. a = m/s?

An object's position as a function of time in one dimension is given by the expression; 3.89t2 + 2.22t + 7.48 where are constants have proper SI Units. What is the object's average velocity between the times t = 3.16 s and t = 8.38 s? with everything in 3 sig figs

You are driving 65.0 miles/hour and you see a police car, so you slow down to 42.5 miles/hour. It takes you 3 seconds to do this. What is your deceleration? (I asked this one before, but someone plugged in -22.5 miles/hr x 2.237 m/s / 1 mile an hour which I didn't understand)

A point particle of mass m = 1.8 kg moves according to the position function: r(t) = xtai + ytbj + ztck, where t denotes time and x, y, z, a, b, and c are constants such that the exponents are positive integers and the position function has the dimension of length.  Part (a)  We can write the particle’s velocity function in the form v(t) = ntdi + otej + ptgk. Enter an expression for n in terms of x, y, z, a, b, and c.   Part (b)  The particle’s velocity function will have the form v(t) = ntdi + otej + ptgk. Enter an expression for d in terms of x, y, z, a, b, and c.  Part (c)  Here is a set of parameter values for the motion of the particle: m = 1.8 kg, x = 1.8 m/s0, y = 2.4 m/s1, z = 0.15 m/s2, a = 0, b = 1, c = 2. Calculate the x-component of the particle’s angular momentum, in units of kg˙m2/s, about the origin at time t = 1 s.   Part (d)  Use the same set of parameter values (m = 1.8 kg, x = 1.8 m/s0, y = 2.4 m/s1, z = 0.15 m/s2, a = 0, b = 1, c = 2) to calculate the y-component of…

A soccer ball starts from rest and accelerates with an acceleration of 0.385 m/s2 while moving down a 8.50 m long inclined plane. When it reaches the bottom, the ball rolls up another plane, where, after moving 14.75 m, it comes to rest. (a) What is the speed of the ball at the bottom of the first plane (in m/s)? (Round your answer to at least two decimal places.) m/s (b) How long does it take to roll down the first plane (in s)? (c) What is the magnitude of the acceleration along the second plane (in m/s?)? m/s2 (d) What is the ball's speed 8.20 m along the second plane (in m/s)? m/s (e) What If? You change the angles of the two inclines in such a manner that the new acceleration down the first incline is a, new down = 2a initiel down and up the second incline is a, new un x initial up, What is the ball's speed 8.20 m up the second plane (in m/s)? 2 m/s

The x component of the velocity of an object moving along the x axis is given as vy = 4t – 3t where vy is in meters per sec and t in seconds. The x component of its average velocity over the interval of time from t =0 to t= 2 sec is: O -2m/s O 2 m/s O -4m/s

A boat running at its top speed takes 3.0 hours to travel 12 km down a river. It then takes 5.0 hours to return to its starting place. What is the velocity of the river? (Answer should look like '0.35 km/h'. I know bad units.)

An object's position in the X-direction as a function of time is given by the expressio%3B x(t) = 5t2 + 2t where are quantities have proper SI Units. What is the object's average velocity in the x-direction between the times t = 1.71 s and t = 2.37 s. Just enter the number rounded to 3 significant figures and assume it has proper SI Units.

v(t) = 4t3 + 10t m/s. The initial position is r = 3 m. 3. A one dimensional problem. Assume v = Determine the position and acceleration at time t = 3sec.