# While driving, one of the tyres of a car hits a nail which causes a change of pressure P(kPa) by 10% every 1 hour. The pressure of the car tyre changes following the relation of P=Poe^(-0.105t) ?

## 1)Before hitting the nail the tyre's initial pressure was 300kPa. Upon reaching 100kPa the car cannot be driven. Determine the duration of the car can be driven before the tyre pressure drops to 100kPA

Jul 4, 2018

$t \approx 10.46 \text{ hrs}$

#### Explanation:

There is no calculus here, either in the derivation of the formula or in the direct question.

• "a change of pressure P(kPa) by [-] 10% every 1 hour."

Each hour:

$P \to 0.9 {P}_{o} \to {0.9}^{2} {P}_{o} \ldots . .$

$\implies P = {P}_{o} {\left(0.9\right)}^{t}$

And

(0.9)^t = e^(ln(0.9)^t) = e^(t ln(0.9)

NB, $\ln \left(0.9\right) \approx - 0.105$

• "Determine the duration of the [time period the] car can be driven before the tyre pressure drops to 100kPA"

$\frac{P}{P} _ o = {e}^{- 0.105 t}$

Take logs:

$- 0.105 t = \ln \left(\frac{P}{P} _ o\right) \left[= - \ln \left({P}_{o} / P\right)\right]$

t =( ln (P_o/P ))/0.105 =( ln (3/1 ))/0.105 approx 10.46" hrs"

Looks like this :)

graph{(y - 3 e^(-0.105 x))(y - 1) = 0 [-1 , 15, -1, 5]}