Question

The position of an object at any time t is given by s(t)=3t4−40t3+126t2−9. i. Determine the velocity of the object at any time t. [5 marks] ii.Does the object ever stop changing? [5 marks] iii.When is the object moving to the right and when is the object moving to the left?

Answer 1

Answer:

1.) V = 12t^3 - 120t^2 + 252t

2.) No

3.) At positive V and negative V

Explanation: Given that the position of an object at any time t is s(t)=3t4−40t3+126t2−9.

1.) To determine the velocity V, we will differenciate the equation above with respect to t.

V = ds/dt

ds/dt = 12t^3 - 120t^2 + 252t

Therefore,

V = 12t^3 - 120t^2 + 252t

2.) The equation above depict an exponential equation which proves that the object never stops changing.

3.) The object moves to the right at positive velocity V and moves to the left at negative velocity V

Answer 2

i. The velocity of the object at any time is V = 12t^3 - 120t^2 + 252t

ii. The object is Not changing.

iii. At positive V and negative V.

Calculation of the velocity:

Since the position of an object at any time t is

s(t)=3t4−40t3+126t2−9.

1.

Here the velocity is

V = ds/dt

ds/dt = 12t^3 - 120t^2 + 252t

SO,

V = 12t^3 - 120t^2 + 252t

2.) The equation above represents an exponential equation that proves that the object never stops changing.

3.) The object moves to the right at positive velocity V and moves to the left at negative velocity V

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