Define (i) velocity, (ii) friction - Leaving Cert Physics - Question 6 - 2009

The distance traveled at top speed can be calculated using the formula:

$s = ut + vt$

Where:

• $u$ = 0 m/s (during acceleration)
• $t$ = 100 seconds (time taken to reach top speed)
• $v$ = 50 m/s (constant speed)

Total time at top speed = 90 minutes = 5400 seconds.

Distance during top speed:

$s = 50 \times 5400 = 270000 \text{ meters}$

Using the formula:

$a = \frac{v - u}{t}$

We need to find the deceleration when brakes were applied. Assuming the final speed ($v$) is 0 m/s at the stop, and distance ($s$) is 500 m:

Using the formula:

$v^2 = u^2 + 2as$

Setting $v = 0$ gives:

$0 = u^2 + 2a(500)\implies a = -\frac{u^2}{1000}$.

Using Newton’s second law:

$F = ma$

Where:

• $m$ = mass = 30000 kg
• $a$ = deceleration found previously.

Plugging in the deceleration will provide the force acting.

The kinetic energy (KE) can be calculated with:

$KE = \frac{1}{2} mv^2$

Substituting values:

$KE = \frac{1}{2} \times 30000 \times (50)^2 = 37500000 \text{ J}$

The kinetic energy lost by the train during stopping is converted into heat, sound, and energy used for braking.

Force A is the driving force (T) from the engines, and Force B is the frictional force opposing the motion.

When force A equals force T, the train moves at a constant velocity, indicating uniform motion.

The graph should show a linear increase to 50 m/s during acceleration, a flat line at 50 m/s during constant speed, and a downward slope back to 0 m/s during deceleration.

• Initial segment: Acceleration phase (straight diagonal line).
• Middle segment: Constant speed (horizontal line).
• Final segment: Deceleration phase (straight diagonal down).