A cyclist is moving along a straight horizontal road and passes a point A. Five seconds later, at the instant when she is moving with speed 10 m/s, she passes the point B. She moves with constant acceleration from A to B. Given that AB = 40m, find (a) the acceleration of the cyclist as she moves from A to B, (b) the time it takes her to travel from A to the midpoint of AB.

Photo AI

A cyclist is moving along a straight horizontal road and passes a point A - Edexcel - A-Level Maths Mechanics - Question 6 - 2017 - Paper 1

Question 6

A cyclist is moving along a straight horizontal road and passes a point A. Five seconds later, at the instant when she is moving with speed 10 m/s, she passes the po... show full transcript

Worked Solution & Example Answer:A cyclist is moving along a straight horizontal road and passes a point A - Edexcel - A-Level Maths Mechanics - Question 6 - 2017 - Paper 1

Step 1

a) the acceleration of the cyclist as she moves from A to B

96%

114 rated

Answer

To find the acceleration, we can use the equation of motion:

$s = v_0 t + \frac{1}{2} a t^2$

Where:

$s = 40$ m (distance from A to B)
$v_0 = 0$ m/s (initial speed at A since she just passed A)
$t = 5$ s (time taken to reach B)

Plugging in the values:

$40 = 0 \cdot 5 + \frac{1}{2} a (5^2)$

This simplifies to:

$40 = \frac{1}{2} a (25)$

Thus,

$40 = 12.5 a$

Solving for $a$ :

$a = \frac{40}{12.5} = 3.2 \, \text{m/s}^2$

Step 2

b) the time it takes her to travel from A to the midpoint of AB

99%

104 rated

Answer

The distance to the midpoint M is:

$s_{AM} = \frac{40}{2} = 20 \, \text{m}$

Using the equation of motion again:

$s = v_0 t + \frac{1}{2} a t^2$

Where now:

$s = 20$ m
$v_0 = 0$ m/s (initial speed at A)
$a = 3.2 \, \text{m/s}^2$

Plugging in the values:

$20 = 0 \cdot t + \frac{1}{2} (3.2) t^2$

This simplifies to:

$20 = 1.6 t^2$

Thus,

$t^2 = \frac{20}{1.6} = 12.5$

Taking the square root gives:

$t = \sqrt{12.5} \approx 3.54 \, \text{s}$

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;