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 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

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 constant ... 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

(b) the value of V

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Answer

To find the value of V, we can consider the distance travelled in the first 20 seconds. Using the equation for uniformly accelerated motion, we have:

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

Where:

S = Distance = 140 m
u = Initial velocity = 0 m/s (from rest)
a = Acceleration, which we can denote as ( a_1 = \frac{V}{20} )
t = time = 20 s.

Substituting in:

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

Solving for V, we get:

140 = \frac{V}{2} \cdot 400 \ \Rightarrow 140 = 200 V \ \Rightarrow V = 0.7 imes 20 = 14 ext{ m/s}.

Step 2

(c) the total time for this journey

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Answer

The total time can be calculated by considering the three phases of the journey:

Acceleration phase: 20 seconds (to reach V)
Constant speed phase: 30 seconds (at speed V)
Deceleration phase:
- From speed 8 m/s to rest with deceleration of 1 m/s²:
- The time taken to decelerate from 8 m/s to rest:
Using (v = u + at): $0 = 8 - 1\cdot t_2 \\ t_2 = 8 ext{ seconds}$

The total time is:

T_{total} = 20 + 30 + 8 = 58 ext{ seconds}.

Step 3

(d) the total distance travelled by the car

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Answer

To find the total distance travelled, we can sum the distances for each phase:

Acceleration phase: Distance is 140 m (calculated previously).
Constant speed phase: The car travels for 30 seconds at 14 m/s:
$D_{constant} = V \cdot 30 = 14 \cdot 30 = 420 ext{ m}.$
Deceleration phase: From 8 m/s to rest with a deceleration of 1 m/s²:

The distance travelled during this phase is given by:
$s = ut + \frac{1}{2}at^2 = 8 \cdot 8 + \frac{1}{2}(-1)(8^2) \ = 64 - 32 = 32 ext{ m}.$

Summing all distances:

D_{total} = 140 + 420 + 32 = 592 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

(b) the value of V

(c) the total time for this journey

(d) the total distance travelled by the car

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