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A car is moving along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 4 - 2007 - Paper 1
Question 4
A car is moving along a straight horizontal road. At time t = 0, the car passes a point A with speed 25 m s⁻¹. The car moves with constant speed 25 m s⁻¹ until t = 1... show full transcript
Worked Solution & Example Answer:A car is moving along a straight horizontal road - Edexcel - A-Level Maths Mechanics - Question 4 - 2007 - Paper 1
Step 1
Sketch the speed-time graph showing the motion of the car from A to B.
Answer
To sketch the speed-time graph, we have two horizontal segments corresponding to the constant speed of 25 m/s from t = 0 to t = 10 s, and then from t = 18 s to t = 30 s. Between t = 10 s to t = 18 s, a downward sloping line represents deceleration as the speed decreases from 25 m/s to V m/s.
Step 2
Find the value of V.
Answer
Using the formula for distance, we calculate the distance covered during each segment:
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From t = 0 to t = 10 s:
Distance = Speed × Time = 25 m/s × 10 s = 250 m
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From t = 10 s to t = 18 s (deceleration): The average speed during this interval is given by
Average Speed = rac{(25 + V)}{2} ext{ m/s}
So, distance from t = 10 s to t = 18 s is:
Distance = ext{Average Speed} × ext{Time} = rac{(25 + V)}{2} × 8 ext{ s}
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From t = 18 s to t = 30 s (constant speed V):
Distance = V × 12 s
The total distance is:
250 + rac{(25 + V)}{2} × 8 + V × 12 = 526
Simplifying:
ightarrow V = 11 ext{ m/s}$$
Step 3
Find the deceleration of the car between t = 10 s and t = 18 s.
Answer
Using the equation of motion:
where:
- v = final speed (V m/s)
- u = initial speed (25 m/s)
- a = acceleration (deceleration in this case)
- t = time interval (8 s)
Substituting the values, we have:
Rearranging gives:
Therefore:
a = - rac{14}{8} = -1.75 ext{ m/s}^2.
Thus, the deceleration of the car is 1.75 m/s².