Question

The position of an object moving along an x axis is given by x = 3.24 t - 4.20 t2 + 1.07 t3, where x is in meters and t in seconds. Find the position of the object at the following values of t: (a) 1 s, (b) 2 s, (c) 3 s, and (d) 4 s. (e) What is the object's displacement between t = 0 and t = 4 s? (f) What is its average velocity from t = 2 s to t = 4 s?

Answer 1

Answer and explanation:

Given : The position of an object moving along an x axis is given by
`x=3.24t-4.20t^2+1.07t^3` where x is in meters and t in seconds.

To find : The position of the object at the following values of t :

a) At t= 1 s

`x(t)=3.24t-4.20t^2+1.07t^3`

`x(1)=3.24(1)-4.20(1)^2+1.07(1)^3`

`x(1)=3.24-4.20+1.07`

`x(1)=0.11`

b) At t= 2 s

`x(t)=3.24t-4.20t^2+1.07t^3`

`x(2)=3.24(2)-4.20(2)^2+1.07(2)^3`

`x(2)=6.48-16.8+8.56`

`x(2)=-1.76`

c) At t= 3 s

`x(t)=3.24t-4.20t^2+1.07t^3`

`x(3)=3.24(3)-4.20(3)^2+1.07(3)^3`

`x(3)=9.72-37.8+28.89`

`x(3)=0.81`

d) At t= 4 s

`x(t)=3.24t-4.20t^2+1.07t^3`

`x(4)=3.24(4)-4.20(4)^2+1.07(4)^3`

`x(4)=12.96-67.2+68.48`

`x(4)=14.24`

(e) What is the object's displacement between t = 0 and t = 4 s?

At t=0, x(0)=0

At t=4, x(4)=14.24

The displacement is given by,

`\triangle x=x(4)-x(0)`

`\triangle x=14.24-0`

`\triangle x=14.24`

(f) What is its average velocity from t = 2 s to t = 4 s?

At t=2, x(2)=-1.76

At t=4, x(4)=14.24

The average velocity is given by,

`\triangle x=x(4)-x(2)`

`\triangle x=14.24-(-1.76)`

`\triangle x=14.24+1.76`

`\triangle x=16`