A car starts from rest and moves with constant acceleration along a straight horizontal road. The car reaches a speed of $V$ m s⁻¹ in 20 seconds. It moves at constant speed $V$ m s⁻¹ for the next 30 seconds, then moves with constant deceleration $1/2$ m s⁻² until it has speed 8 m s⁻¹. It moves at speed 8 m s⁻¹ for the next 15 seconds and then moves with constant deceleration $1/3$ m s⁻² until it comes to rest. In the first 20 seconds of this journey the car travels 140 m. Find (b) the value of $V$; (c) the total time for this journey; (d) the total distance travelled by the car.

Photo AI

A car starts from rest and moves with constant acceleration along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 3 - 2014 - Paper 1

Question 3

A car starts from rest and moves with constant acceleration along a straight horizontal road. The car reaches a speed of $V$ m s⁻¹ in 20 seconds. It moves at constan... show full transcript

Worked Solution & Example Answer:A car starts from rest and moves with constant acceleration along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 3 - 2014 - Paper 1

Step 1

Find the value of $V$

96%

114 rated

Answer

Using the kinematic equation for distance under constant acceleration, we have:

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

The initial velocity $u = 0$ , and the time $t = 20$ s. We also have:

$140 = 0 + \frac{1}{2} a (20)^2$

This equation can be rearranged to solve for $a$ :

$a = \frac{140 imes 2}{20^2} = 0.7 \text{ m s}^{-2}$

Now, we can find the speed at 20 seconds:

$V = u + at = 0 + 0.7 \times 20 = 14 \text{ m s}^{-1}$

Thus, the value of $V$ is $14$ .

Step 2

Find the total time for this journey

99%

104 rated

Answer

We'll need to calculate the duration for each segment of the journey:

First segment (0-20 seconds):
- Duration: 20 seconds
Second segment (20-50 seconds):
- Speed: $V = 14$ m/s and time: 30 seconds.
- Duration: 30 seconds
Third segment (50-65 seconds):
- Initial speed: $14$ m/s, final speed: $8$ m/s; deceleration: $0.5$ m/s².
- Using $v = u + at$ :
- Solving gives $t = \frac{14 - 8}{0.5} = 12$ seconds.
Fourth segment (65-80 seconds):
- Speed: 8 m/s, constant for 15 seconds.
- Duration: 15 seconds
Fifth segment (deceleration to rest from speed 8):
- Deceleration: $1/3$ m/s². Using the formula $v = u + at$ , we find:
- Time taken to stop:
- $t = \frac{8}{1/3} = 24$ seconds.

Total time:
$T = 20 + 30 + 12 + 15 + 24 = 101 ext{ seconds}$

Step 3

Find the total distance travelled by the car

96%

101 rated

Answer

Total distance ( $D$ ) can be calculated by summing the distances of all segments:

First segment: $D_1 = 140 ext{ m}$
Second segment: $D_2 = V \times 30 = 14 \times 30 = 420 ext{ m}$
Third segment (decelerating from 14 m/s to 8 m/s):
- Average speed during this period: $\text{Average speed} = \frac{14 + 8}{2} = 11 ext{ m/s}$
- Distance: $D_3 = average imes time = 11 \times 12 = 132 ext{ m}$
Fourth segment (constant speed at 8 m/s): $D_4 = 8 \times 15 = 120 ext{ m}$
Fifth segment (decelerating from 8 m/s to rest):
- Average speed: $\text{Average speed} = \frac{8 + 0}{2} = 4 ext{ m/s}$
- Distance: $D_5 = 4 \times 24 = 96 ext{ m}$

Total distance:
$D = D_1 + D_2 + D_3 + D_4 + D_5 = 140 + 420 + 132 + 120 + 96 = 908 ext{ m}$

A car starts from rest and moves with constant acceleration along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 3 - 2014 - Paper 1

Worked Solution & Example Answer:A car starts from rest and moves with constant acceleration along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 3 - 2014 - Paper 1

Find the value of $V$

Find the total time for this journey

Find the total distance travelled by the car

Join the A-Level students using SimpleStudy...