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.

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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

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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 co... 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

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Answer

Using the kinematic equation for constant acceleration:

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

Here, initial velocity $u = 0$ , acceleration $a = \frac{V}{20}$ , and time $t = 20$ seconds. Thus, the distance $s$ for the first 20 seconds is:

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

This simplifies to:

$140 = \frac{V}{20} \cdot 200$

From this, we can find:

V = 14 ext{ m/s}$$

Step 2

Find the total time for this journey

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Answer

First, we find the time taken to decelerate from $V$ m/s to 8 m/s. Using the equation:

$v = u + at$

Where $v = 8$ m/s, $u = 14$ m/s, and $a = -\frac{1}{2}$ m/s $^2$ . Therefore:

t_{1} = 12 ext{ seconds}$$ Next, we find the time taken to decelerate from 8 m/s to rest with an acceleration of $-\frac{1}{3}$ m/s$^2$: Using the same formula: $$0 = 8 - \frac{1}{3}t_{2} \Rightarrow \frac{1}{3}t_{2} = 8 \\ t_{2} = 24 ext{ seconds}$$ Adding up all the times, we get: $$t_{total} = 20 + 30 + t_{1} + t_{2} = 20 + 30 + 12 + 24 = 86 ext{ seconds}$$

Step 3

Find the total distance travelled by the car

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Answer

To calculate the total distance travelled, we can sum the distances for each segment:

Acceleration phase (first 20 seconds):
- Distance = 140 m
Constant speed phase (next 30 seconds):
- Distance = $V \times 30 = 14 \times 30 = 420$ m
Deceleration to 8 m/s: Using the formula: $s = ut + \frac{1}{2}at^2$
- Initial speed = 14 m/s, final speed = 8 m/s, time = 12 s:
- Average speed = $\frac{14 + 8}{2} = 11$ m/s.
- Distance = $11 \times 12 = 132$ m
Constant speed phase (15 seconds at 8 m/s):
- Distance = $8 \times 15 = 120$ m
Deceleration to rest:
- Initial speed = 8 m/s, final speed = 0 m/s, time = 24 s:
- Average speed = $\frac{8 + 0}{2} = 4$ m/s.
- Distance = $4 \times 24 = 96$ m

Now, summing these distances gives:

$140 + 420 + 132 + 120 + 96 = 908 ext{ m}$

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

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