The following kinematics formulas may be used in this question. v = u + at s = ut + \frac{1}{2}at^2 v^2 = u^2 + 2as The initial velocity of a particle is 20 m/s. The acceleration of the particle is 8 m/s². After t seconds, the particle has travelled 25 m. a) Show that 4t² - 20t + 25 = 0. b) Solve 4t² - 20t + 25 = 0. c) Show that the particle is stationary when it has travelled 25 m.

Photo AI

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

Question 16

The following kinematics formulas may be used in this question. v = u + at s = ut + \frac{1}{2}at^2 v^2 = u^2 + 2as The initial velocity of a particle is 20 m/s. Th... show full transcript

Worked Solution & Example Answer:The following kinematics formulas may be used in this question - OCR - GCSE Maths - Question 16 - 2021 - Paper 1

Step 1

a) Show that 4t² - 20t + 25 = 0.

96%

114 rated

Answer

To show this, we can use the kinematic equation for displacement:

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

Substituting the given values:

Initial velocity, $u = 20$ m/s
Acceleration, $a = 8$ m/s²
Displacement, $s = 25$ m

This gives us:

$25 = 20t + \frac{1}{2}(8)t^2$

Simplifying the equation:

$25 = 20t + 4t^2$

Rearranging this, we have:

$4t^2 - 20t + 25 = 0$

Thus, we have shown the required equation.

Step 2

b) Solve 4t² - 20t + 25 = 0.

99%

104 rated

Answer

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:

$t = 2.5$

Step 3

c) Show that the particle is stationary when it has travelled 25 m.

96%

101 rated

Answer

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.

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

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

Worked Solution & Example Answer:The following kinematics formulas may be used in this question - OCR - GCSE Maths - Question 16 - 2021 - Paper 1

a) Show that 4t² - 20t + 25 = 0.

b) Solve 4t² - 20t + 25 = 0.

c) Show that the particle is stationary when it has travelled 25 m.

Join the GCSE students using SimpleStudy...

Other GCSE Maths topics to explore