Photo AI
A car moves along a straight horizontal road from a point A to a point B, where AB = 885 m - Edexcel - A-Level Maths: Mechanics - Question 6 - 2012 - Paper 1
Question 6
A car moves along a straight horizontal road from a point A to a point B, where AB = 885 m. The car accelerates from rest at A to a speed of 15 m s⁻¹ at a constant r... show full transcript
Worked Solution & Example Answer:A car moves along a straight horizontal road from a point A to a point B, where AB = 885 m - Edexcel - A-Level Maths: Mechanics - Question 6 - 2012 - Paper 1
Step 1
Find the time for which the car decelerates.
Answer
To find the time the car decelerates, we first need to determine the entire time of the journey.
-
Acceleration Phase:
- Initial speed, ( u = 0 , \text{m s}^{-1} )
- Final speed, ( v = 15 , \text{m s}^{-1} )
- Acceleration time, ( t_a = \frac{1}{3} T )
- Using the equation, ( v = u + at ):
[ 15 = 0 + a \cdot \frac{1}{3} T \Rightarrow a = \frac{45}{T} ]
-
Constant Speed Phase:
- Time at constant speed, ( t_c = T , \text{seconds} )
-
Deceleration Phase:
- Final speed at B, ( v = 0 , \text{m s}^{-1} )
- Deceleration, ( a_d = -2.5 , \text{m s}^{-2} )
- Distance covered while decelerating, use displacement:
[ rac{1}{2} \cdot (15 + 0) \cdot t_d = \text{Distance during deceleration} ]
We need the total distance:
- Distance during acceleration: ( \frac{1}{2} \cdot u \cdot t_a + \frac{1}{2} a t_a^2 = \frac{1}{2} \cdot 15 , \text{m s}^{-1} , \cdot \left( \frac{1}{3} T \right) = \frac{5T}{6} )
- Distance during constant speed: ( 15 \cdot T )
- Total distance: [ \frac{5T}{6} + 15T + \frac{1}{2} \cdot 15 \cdot t_d = 885 ]
Setting up the equation for deceleration:
Solve the above to find ( t_d ) and hence the deceleration time.
Step 2
Sketch a speed-time graph for the motion of the car.
Answer
-
Graph Segments:
- The first segment rises from (0,0) to ( t_a = \frac{1}{3}T ) at (( \frac{1}{3}T, 15 )).
- The second segment is a horizontal line from ( t_a ) to ( t_a + T ).
- The third segment descends back to (Total Time, 0) at ( t_a + T + t_d ).
-
Shape and Labels:
- Make sure to clearly label the axes: vertical for speed in m/s and horizontal for time in seconds, with all relevant points marked.
Step 3
Step 4
Step 5
Sketch an acceleration-time graph for the motion of the car.
Answer
-
Graph Segments:
- Start from 0 m/s² during acceleration, rise to ( \frac{45}{48} ) until time ( t_a ).
- Maintain constant acceleration until ( t_a + T ).
- Then drop to -2.5 during deceleration.
-
Axes and Labels:
- Clearly label the vertical axis for acceleration in m/s² and the horizontal axis for time in seconds, marking all relevant points.