The following kinematics formulas may be used in this question - OCR - GCSE Maths - Question 16 - 2021 - Paper 1

This quadratic equation can be solved using the quadratic formula:

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

Here, $a = 4$, $b = -20$, and $c = 25$. Substituting these values:

• Calculate the discriminant:

$D = b^2 - 4ac = (-20)^2 - 4(4)(25) = 400 - 400 = 0$

Since the discriminant is zero, there is one repeated solution:

$t = \frac{20 \pm \sqrt{0}}{2 \times 4} = \frac{20}{8} = 2.5$

Thus, the solution is:

$t = 2.5$

A particle is stationary when its velocity is zero. Using the equation:

$v = u + at$

Substituting the values:

• Initial velocity, $u = 20$ m/s
• Acceleration, $a = 8$ m/s²
• We have already found time, $t = 2.5$ s.

Now substituting for $t$:

$v = 20 + 8(2.5) = 20 + 20 = 40 \, \text{m/s}$

We need to assess if the particle was stationary during its travel. The particle stops being stationary after the displacement of 25 m, which confirms it has a velocity of 0 only at the start. Hence, the motion equation concludes that it was stationary just as it started.