A motorist, travelling at a constant velocity of V m/s, passed a fixed point X and saw few vehicles ahead. He stepped on the accelerator and his subsequent velocity, v m/s, is given by $v = 60e^{t/6}$, where $t$ is the time in seconds after passing X. As he passed a point Y, his velocity has increased to twice his velocity at X.

1. Find the initial velocity at point X

At point X, $t = 0$.
$V = 60e^{0/6} = 60\text{ m/s}$

2. Set up the equation for the time at point Y

Velocity at Y is twice the velocity at X, so $v = 2 \times 60 = 120\text{ m/s}$.
$120 = 60e^{t/6}$

3. Solve the exponential equation for t

$e^{t/6} = 2$
$\frac{t}{6} = \ln 2$
$t = 6\ln 2 \approx 4.15888 \approx 4.16\text{ s}$

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

$a = \frac{dv}{dt} = \frac{d}{dt}(60e^{t/6}) = 60 \left(\frac{1}{6}\right) e^{t/6} = 10e^{t/6}$

2. Evaluate the acceleration at point Y.

At Y, the velocity is twice the velocity at X ($v = 120$), which means $60e^{t/6} = 120 \implies e^{t/6} = 2$.
Substitute this into the acceleration expression:
$a = 10(2) = 20$

1. Set up the definite integral for the distance travelled from X to Y

From part (a), the time at X is $t = 0$ and the time at Y is $t = 6\ln 2$.
$\text{Distance } XY = \int_{0}^{6\ln 2} 60e^{t/6} \, dt$

2. Integrate the velocity function

$\int 60e^{t/6} \, dt = \frac{60}{1/6}e^{t/6} = 360e^{t/6}$

3. Evaluate the definite integral to find the distance

$\text{Distance } XY = \left[ 360e^{t/6} \right]_{0}^{6\ln 2}$
$= 360e^{\frac{6\ln 2}{6}} - 360e^{0}$
$= 360e^{\ln 2} - 360(1)$
$= 360(2) - 360 = 360$