A car moves on a straight road such that its position from a reference point is

Question

A car moves on a straight road such that its position from a reference point is given by the following function: {x(t) = 3 + 5t + 2t^2 + 0.4t^3, y(t) = 0, z(t) = 0} where t is in seconds and x in meters. True or False: the acceleration of the car is constant? Find the acceleration, in meters per square second, in the x-direction at t_1 = 0.75 sec. Find the acceleration, in meters per square second, in the x-direction at t = 1.7 sec. Find the average velocity, in meters per second, in the x-direction between t_1 = 0.75 s and t_2 = 1.7 s. Find the average acceleration, in meters per second squared, in the x-direction between t_1 and t_2 = 1.7 s.

Explanation / Answer

x(t) = 3 + 5t + 2t^2 + 0.4t^3

velocity = dx(t)/dt

velocity = 5 + 4t + 1.2t^2

acceleration = dvelocity / dt

acceleration = 4 + 2.4t

since acceleration id depending upon the time so its not constant

when t = 0.75 sec

acceleration = 4 + 2.4 * 0.75

acceleration = 5.8 m/s^2

when t = 1.7 sec

acceleration = 4 + 2.4 * 1.7

acceleration = 8.08 m/s^2

average velocity = total distance / total time

total distance = x(1.7) - x(0.75)

average velocity = (3 + 5 * 1.7 + 2 * 1.7^2 + 0.4 * 1.7^3) - (1 + 5 * 0.75 + 2 * 0.75^2 + 0.4 * 0.75^3) / (1.7 - 0.75)

average velocity = 13.896 m/s

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