A sports car driven along a straight road passes a traffic junction $A$ at $p$ m/s. A little later, it passes a second traffic junction $B$ with a speed of $40$ m/s. Between $A$ and $B$, the speed of the car, $v$ m/s, is given by $v = 5e^{0.05t} + 10$ where $t$ is the time in seconds after passing $A$.

1. Substitute $t = 0$ into the velocity equation to find the initial speed

At junction $A$, $t = 0$.
$v = 5e^{0.05(0)} + 10$

2. Evaluate the expression to find $p$

$v = 5(1) + 10 = 15$
$\therefore p = 15$

1. Set up the equation for the time at junction $B$

At junction $B$, $v = 40$.
$5e^{0.05t} + 10 = 40$

2. Isolate the exponential term

$5e^{0.05t} = 30$
$e^{0.05t} = 6$

3. Solve for $t$ using natural logarithms

$0.05t = \ln 6$
$t = \frac{\ln 6}{0.05}$
$t = 20 \ln 6$ (Shown)

1. Set up the definite integral for the distance $AB$.

Distance $AB = \int_{0}^{20 \ln 6} v \, dt$
$= \int_{0}^{20 \ln 6} (5e^{0.05t} + 10) \, dt$

2. Integrate the velocity function.

$\int (5e^{0.05t} + 10) \, dt = \left[ \frac{5}{0.05}e^{0.05t} + 10t \right]_{0}^{20 \ln 6}$
$= \left[ 100e^{0.05t} + 10t \right]_{0}^{20 \ln 6}$

3. Evaluate the definite integral using the limits.

Substitute $t = 20 \ln 6$:
$100e^{0.05(20 \ln 6)} + 10(20 \ln 6) = 100e^{\ln 6} + 200 \ln 6 = 100(6) + 200 \ln 6 = 600 + 200 \ln 6$

Substitute $t = 0$:
$100e^{0} + 10(0) = 100$

Distance $= (600 + 200 \ln 6) - 100$
$= 500 + 200 \ln 6$

4. Calculate the final numerical value.

$500 + 200 \ln 6 \approx 858.35 \approx 858$ m (to 3 s.f.)

1. Find the expression for acceleration by differentiating velocity with respect to time.

$a = \frac{dv}{dt} = \frac{d}{dt}(5e^{0.05t} + 10)$
$a = 5(0.05)e^{0.05t} = 0.25e^{0.05t}$

2. Substitute $t = 30$ to find the acceleration.

$a = 0.25e^{0.05(30)}$
$a = 0.25e^{1.5}$
$a \approx 1.1204 \text{ m/s}^2$
$a \approx 1.12 \text{ m/s}^2 \text{ (to 3 s.f.)}$