Question

The position of a particle moving along the x axis depends on the time according to the equation x = ct² - bt³, where x is in meters and t in seconds.

For the following, let the numerical values of a and b be 3.7 and 1.0 respectively.

a) At what time does the particle reach its maximum positive x position?

b) What distance does the particle cover in the first 4.0s?

c) What is its displacement from t=0s to t-4.0s?

d) What is its velocity at t = 1.0?

What is its velocity at t = 2.0?

What is its velocity at t = 3.0?

What is its velocity at t = 4.0 s?


e) What is its acceleration at t = 1.0 s?
m/s²
What is its acceleration at t = 2.0 s?
m/s²
What is its acceleration at t = 3.0 s?
m/s²
What is its acceleration at t = 4.0 s?
m/s²

Answer 1

The position of particel as a function of time is given by

x=3.7t²-t³

a) for x to be maximum, dx/dt=0

dx/dt = 7.4t-3t² = 0

t=0 or t= 2.467 second

x(0)=3.7*0²-0³=0 and x(2.467)=3.7*2.467²-2.467³=7.5 m

x reached its maximum positive value at 2.467 seconds

b) distnace travelled by particle in time 0 to 4 seconds is given by

because distnace is always positive.

between 0 and 2.467m, the velocity is positive and beyond 2.467 m, the velocity is negative

distnace travelled =

c) displacement is given by

Displacement = -4.8 m

d) velocity = dx/dt = 7.4t-3t²

velocity at t=1 = 7.4*1-3*1² = 4.4 m/s

velocity at t=2 = 7.4*2-3*2² = 2.8 m/s

velocity at t=3 = 7.4*3-3*3² = -4.8 m/s

velocity at t=4 = 7.4*4-3*4² = -18.4 m/s

e) acceleration = dv/dt = 7.4-6t

acceleration at t=1 = 7.4-6*1=1.4 m/s²

acceleration at t=2 = 7.4-6*2=-4.6 m/s²

acceleration at t=3 = 7.4-6*3=-10.6 m/s²

acceleration at t=4 = 7.4-6*4=-16.6 m/s²