Question

The position of a particle moving along the x axis is given in centimeters by x = 9.16 + 1.87 t3, where t is in seconds. Calculate (a) the average velocity during the time interval t = 2.00 s to t = 3.00 s; (b) the instantaneous velocity at t = 2.00 s; (c) the instantaneous velocity at t = 3.00 s; (d) the instantaneous velocity at t = 2.50 s; and (e) the instantaneous velocity when the particle is midway between its positions at t = 2.00 s and t = 3.00 s.

Answer 1

a)V=17.76 cm/s

b)V=22.44 cm/s

c)V=50.49 cm/s

d)V=35.06 cm/s

e)V=37.63 cm/s

Step-by-step explanation:

Given that

`x=9.16+1.87t^3`

a)

When t= 2 s

`x_1=9.16+1.87* 2^3`

`x_1=24.12\ cm`

When t= 3 s

`x_2=9.16+1.87* 3^3`

`x_2=59.65\ cm`

So the average velocity V

`V=(x_2-x_1 )/(t_2-t_1)`

`V=(59.65- 24.12)/(3-1)`

V=17.76 cm/s

b)

When know that

`V=(dx )/(dt)`

So

`(dx )/(dt)=3* 1.87t^2`

When t= 2 s

`V=3* 1.87* 2^2`

V=22.44 cm/s

c)

At t= 3 s

`V=3* 1.87* 3^2`

V=50.49 cm/s

d)

At t= 2.5 s

`V=3* 1.87* 2.5^2`

V=35.06 cm/s

e)

At t= 2 s ,x=24.12 cm

At t=3 s x=59.65 cm

So midway of 24.12 cm and 59.65 cm is (24.12+59.65 )/2=41.88 cm

So when x=41.88 cm

`x=9.16+1.87t^3`

`41.88=9.16+1.87t^3`

t=2.59 s

So velocity

`V=3* 1.87t^2`

`V=3* 1.87* 2.59^2`

V=37.63 cm/s