A train moves along a straight track with constant acceleration - Edexcel - A-Level Maths Mechanics - Question 3 - 2006 - Paper 1

We can use the equation:

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

where:

• $u = 22.5 \, \text{ms}^{-1}$ (speed at B),
• $a = 2.5 \, \text{ms}^{-2}$ (calculated from part a), and
• $s = 50 \, \text{m}$ (distance to C).

Substituting the values:

$v^2 = (22.5)^2 + 2 \times 2.5 \times 50$

Calculating gives:

$v^2 = 506.25 + 250 = 756.25$

Taking the square root results in:

$v = \sqrt{756.25} \approx 27.5 \, \text{ms}^{-1}$

Using the formula for time:

$s = ut + \frac{1}{2} a t^2$

where:

• $s = 50 \, \text{m}$,
• $u = 22.5 \, \text{ms}^{-1}$,
• $a = 2.5 \, \text{ms}^{-2}$.

Substituting the values:

$50 = 22.5t + \frac{1}{2}(2.5)t^2$

This simplifies to:

$50 = 22.5t + 1.25t^2$

Rearranging leads to:

$1.25t^2 + 22.5t - 50 = 0$

Using the quadratic formula:

$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 1.25$, $b = 22.5$, and $c = -50$. Calculating:

$t = \frac{-22.5 \pm \sqrt{(22.5)^2 - 4 \times 1.25 \times (-50)}}{2 \times 1.25}$

Solving gives:

$t \approx 1.69 \, \text{s}$