Question

A particle of mass m=5.00 kilograms is at rest at t=0.00 seconds. a varying force f(t)=6.00t2−4.00t+3.00 is acting on the particle between t=0.00 seconds and t=5.00 seconds. find the speed v of the particle at t=5.00 seconds. express your answer in meters per second to three significant figures.

Answer 1

The speed v of the particle at t=5.00 seconds = 43 m/s

Step-by-step explanation:

Given :

mass m = 5.00 kg

force f(t) = 6.00t2−4.00t+3.00 N

time t between t=0.00 seconds and t=5.00 seconds

From mathematical expression of Newton's second law;

Force = mass (m) x acceleration (a)

F = ma

`a = (F)/(m)` ...... (1)

acceleration (a)
`= (dv)/(dt)` ......(2)

substituting (2) into (1)

Hence, F
`= (mdv)/(dt)`

`(dv)/(dt) = (F)/(m)`

`dv = (F)/(m) dt`

`dv = (1)/(m)Fdt`

Integrating both sides

`\int\limits {} \, dv = (1)/(m) \int\limits {F(t)} \, dt`

The force is acting on the particle between t=0.00 seconds and t=5.00 seconds;

`v = (1)/(m) \int\limits^5_0 {F(t)} \, dt` ......(3)

Substituting the mass (m) =5.00 kg of the particle, equation of the varying force f(t)=6.00t2−4.00t+3.00 and calculating speed at t = 5.00seconds into (3):

`v = (1)/(5) \int\limits^5_0 {(6t^(2) - 4t + 3)} \, dt`

`v = (1)/(5) |(6t^(3) )/(3) - (4t^(2) )/(2) + 3t |^(5)_(0)`

`v = (1)/(5) |((6(5)^(3) )/(3) - (4(5)^(2) )/(2) + 3(5)) - 0|`

`v = (1)/(5) |(6(125))/(3) - (4(25))/(2) + 15 |`

`v = (1)/(5) |(750)/(3) - (100)/(2) + 15 |`

`v = (1)/(5) | 250 - 50 + 15 |`

`v = (215)/(5)`

v = 43 meters per second

The speed v of the particle at t=5.00 seconds = 43 m/s