# The time that it will take to travel 300 ft. if the initial velocity is 20ft/second and final velocity is 60ft/second.

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Since the velocity is not constant, solve for the average velocity.

`bar(v)=(v_f+v_i)/2`

`bar(v)=(60+20)/2=80/2=40`

Then, apply the formula:

`bar(v)=d/t`

where d - displacement and t - total time.

So,

`40=300/t`

`40t=300`

`t=300/40`

`t=7.5`

**Hence, it took 7.5 seconds to travel 300ft with the given velocities.**

the equation of motion are:

`s=s_0+at` and : `T=T_0+s_0t +1/2 at^2`

where s is speed, a acceleration and T the space.

(`s_0` and`T_0` are the initial speed and the initial sapce forom a system of coordinates. We further supposed the uniformly accelerated motion with a=costant)

so we have:

`at= s-s_0` (1)

`T-T_0=s_0t+1/2at^2` (2)

We know both value `s-s_0= 60-20=40` than `T-T_0=300`

that is: `at=40` (3)

Substituing this in (2):

`T-T_0= s_0t+ 20t`

thus:

`300=20t+20t=40t` `t= 7,5sec`