# If the initial velocity of a car is 6 m/s and the acceleration is related to velocity by a(v) = 12v - 6, what is the distance traveled in the first 3s.

thilina-g | College Teacher | (Level 1) Educator

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The given equation is `a(v) = 12v-6` and initial conditions are v=6 at t=0 and s = 0 when t = 0.

`a = 12v -6`

We can write acceleration as,  `a = (dv)/(dt)`

Therefore,

`(dv)/(dt) = 12v-6`

Separating variables would give,

`(dv)/(2v-1) = 6dt`

By integrating,

`int1/(2v-1)dv = int6dt`

`1/2 ln(2v-1) = 6t + c`

at t = 0, v = 6, then,

`1/2ln(12-1) = 0+c`

`c = 1/2ln(11)`

Therefore the equation is,

`1/2 ln(2v-1) = 6t + 1/2ln(11)`

We can rearrange ln terms together,

`6t = 1/2ln((2v-1)/11)`

`ln((2v-1)/11) = 12t`

Taking the exponential,

`(2v-1)/11 = e^(12t)`

`2v-1 = 11e^(12t)`

`v = 1/2+11/2e^(12t)`

This is the velocity profile of the car. We can find the distance by using this velocity profile. We have to find the distance travelled, s.

We know, `s = intvdt`

Therefore the distance travelled in 3 seconds is,

`s = int_0^3vdt`

`s = int_0^3(1/2+11/2e^(12t))dt`

`int(1/2+11/2e^(12t))dt = 1/2t+11/2e^(12t)/12`

`int(1/2+11/2e^(12t))dt = 1/2t+11/24e^(12t)`

Therefore,

`s = int_0^3(1/2+11/2e^(12t))dt = (3/2+11/24e^(36)-(0+11/24))`

`s = 3/2-11/24+11/24e^36`

`s = 25/24+11/24e^36`

Therefore the distance travelled in 3 seconds is, `s = 25/24+11/24e^36`