A car is moving along a straight horizontal road. The speed of the car as it passes the point A is 25 m s⁻¹ and the car maintains this speed for 30 s. The car then decelerates uniformly to a speed of 10 m s⁻¹. The speed of 10 m s⁻¹ is then maintained until the car passes the point B. The time taken to travel from A to B is 90 s and AB = 1410 m. (a) Sketch, in the space below, a speed-time graph to show the motion of the car from A to B. (b) Calculate the deceleration of the car as it decelerates from 25 m s⁻¹ to 10 m s⁻¹.

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A car is moving along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 4 - 2008 - Paper 1

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A car is moving along a straight horizontal road. The speed of the car as it passes the point A is 25 m s⁻¹ and the car maintains this speed for 30 s. The car then d... show full transcript

Worked Solution & Example Answer:A car is moving along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 4 - 2008 - Paper 1

Step 1

Sketch a speed-time graph

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Answer

To sketch the speed-time graph:

Draw a vertical axis (speed in m s⁻¹) and a horizontal axis (time in seconds).
At time t = 0 s, mark a point at 25 m s⁻¹.
From 0 to 30 s, draw a horizontal line at 25 m s⁻¹ (constant speed).
From 30 s to a time when the speed decreases to 10 m s⁻¹, the line slopes down. The deceleration occurs over 8 seconds, which can be calculated based on the total travel time of 90 s.
From the point where speed reaches 10 m s⁻¹, draw a horizontal line until t = 90 s.

Step 2

Calculate the deceleration of the car as it decelerates from 25 m s⁻¹ to 10 m s⁻¹

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Answer

To calculate the deceleration, use the formula:

$a = \frac{v_f - v_i}{t}$

Where:

$v_f = 10\, \text{m s}^{-1}$ (final speed)
$v_i = 25\, \text{m s}^{-1}$ (initial speed)
$t = 8\, \text{s}$ (time taken to decelerate)

Replacing the values:

$a = \frac{10 - 25}{8} = \frac{-15}{8} = -1.875\, \text{m s}^{-2}$

Thus, the deceleration of the car is $1.875\, \text{m s}^{-2}$ .

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