Question

An object with a mass 4.15kg undergoes an acceleration of 2.7 m/s^2 for a period of 8.0s. It then undergoes an acceleration of -3.2m/s^2 for a period of 20.0s. If the initial velocity of the object is 6.01 m/s and it first expereinces the acceleration when it is at the origin, plot by hand on graph paper the force, velocity, and position as functions of time. Explain why your graphs look the way they do. Include all relevant equations.

Answer 1

Acceleration for the time t= 0 to t=8 is a = 2.7 m/s²
Hence the force for this time , F = ma = 4.15*2.7 = 11.205 N
Now the acceleration has been changed to a = -3.2 m/s² , therefore
the force for rest of time
F = ma = -4.15*3.2 = - 13.208 N
hence the force will remian constant for the period , t= 0 to t =8 and it's magnitude will be F = 11.205 N
and it will be -13.208 N for t = 8 s to t = 28 s
garph will be as shown in the fig.
Now for velocity
first t = 0 t =8
at t =0 , u = 6.01 m/s
and t =8
V = u+at = 6.01 +(2.7*8) = 27.61 m/s
Now from t = 8 to t = 28 s
V_f = 27.61 - (3.2*20) = -36.39 m/s
Now for the distance
Initial at the origin (X_o) = 0
X = X_o+(ut)+(1/2)at²
For t =0 to t = 8 s
X = 0 +(6.01*8) +(1/2)*(2.7)*8² = 134.48 m