Question

An object of mass 4.15 kg undergoes an acceleration of 3.5 m/s^2 for a period of 7.0 seconds. It then undergoes an acceleration of -4.2 m/s^2 for a period of 18.0 seconds. If the initial velocity of the object is 4.01 m/s and it first experiences the acceleration when it is at the origin, plot by hand on graph paper (to scale) the force, velocity, and position graphs as functions of time. Explain why your graphs look the way they do and include all relevant equations.

Answer 1

For Force - Time graph ;

Force is directly proportional to the mass and acceleration

and force , F = m*a , Since in our case mass is constant for entire time period and mass m = 4.15 kg and so, force will varry according to the value of acceleration

a) for period of first 7 second, the acceleration is constant so force will be also constant

acceleration, a = 3.5 m/s²

so, Force F = 4.15 * 3.5 = 14.525 N

b) for period of next 18 second, after changed value of acceleration (a = -4.2 m/s²)it remains constant so, force will also remain constant  

F = 4.15 * (-4.2) = -17.43 N

b) Velocity time graph:

Initial velocty, u = 4.01 m/s

for first 7 second, acceleration a = 3.5 m/s² since the body is accelerating at constant value so, value of velocity will increase linearly as a = dV / dt and intigration of constant value gives linear graph

so, velocity at the end of 7 second can be find by using

first laws of motion, law's of motion as :

V₇ = u + a*t₁

= 4.01 + 3.5 * 7 = 28.51 m/s

for next 18 second, body is decelerating at constant value so, velocity will decrease linearly from V₇

now calculate time required to become velocity equal to zero,

V = u + a*t

0 = 28.51 - 4.2 * t

t = 6.78 second

so, net time = 7 + 6.78 = 13.78 second, aafter this time velocity of the body will have negative velocity

now velocity at the end of 25 second will be as

V₂₅ = V₇ + a₂ * t₂

= 28.51 - 4.2* 18 = -47.09 second

For position time graph :

position, a = dv/ dt = d²s / dt²

and integratin of linear graph(velocity) gives quadratic graph

for first seven second,

sice initialy body at the origin and velocity is increasing costantly , so there will be positive displacement of the object , so the graph will be of increasing nature with time as according to Second laws of motion

s = u*t + 1/2 *a*t² (follows equation of parabola)

and positon of the body after seven seconds

s₇ = 4.01 * 7 + 1/2 * 3.5 * 7² = 113.82 m

Now after seven seconds velocity of the body is decreasing constantly, so, the displacement will start decreasing with time

first of all find time at which the displacement of the object will be zero with the action of deceleration of the mass

s = u*t + 1/2 * a₂*t²

0 = 28.51*t -0.5*4.2*t² (here initial velocity u will be the velocity of body at t = 7 second)

2.1t = 28.51

so, t = 13.58 second

so, Actual time at which displacement of the body is reduced to zero is at (t = 13.58 + 7 = 20.58 second ) after this time there will be negative displacement of the body

to find final displacement of the mass

S = u*t + 1/2 *a *t² + S₇(displacement at the end of seventh second)

here, u = V₇ = 28.51 m/s, t = 18 second and a = - 4.2 m/s²

so, S = 28.51 * 18 - 0.5*4.2*18² + s₇  =113.82 -167.22 m

só, final position of the object. S_{25 second} = -167.22 + 113.82 = -53.4 m