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

We can use the quadratic formula to solve $4t^2 - 20t + 25 = 0$. The quadratic formula is:

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

For our equation:

• $a = 4$
• $b = -20$
• $c = 25$

Substituting these values into the quadratic formula:

$t = \frac{-(-20) \pm \sqrt{(-20)^2 - 4(4)(25)}}{2(4)}$

Calculating:

1. Calculate the discriminant: $(-20)^2 - 4(4)(25) = 400 - 400 = 0$ The discriminant is 0, indicating one real solution.

2. Substitute back to find $t$: $t = \frac{20 \pm \sqrt{0}}{8}$ $t = \frac{20}{8} = 2.5 \text{ s}$

Thus, the solution is: $t = 2.5 \, \text{s}$

To determine when the particle is stationary, we need to find the time when its velocity becomes zero. From the kinematic equation:

$v = u + at$

Substituting known values:

• Initial velocity, $u = 20 \, \text{m/s}$
• Acceleration, $a = 8 \, \text{m/s}^2$
• Let $t$ be the time when velocity is zero.

Setting the equation to zero:

$0 = 20 + 8t$

Rearranging gives:

$8t = -20 \Rightarrow t = -\frac{20}{8} = -2.5 \, \text{s}$

Since time cannot be negative, we need to reconsider if the time spent travelling 25 m leads to a point of rest. From previous calculations, we found that when $t = 2.5$ s, the distance travelled is indeed 25 m. Thus, this is when the particle is stationary at that distance.