A car accelerates uniformly from rest for 20 seconds - Edexcel - A-Level Maths Mechanics - Question 5 - 2011 - Paper 1

To sketch the acceleration-time graph:

1. From 0 to 20 seconds, the car accelerates uniformly, so draw a horizontal line above the time axis, indicating constant positive acceleration.
2. From 20 to 60 seconds, since the car moves at constant speed, draw a line along the time axis (indicating zero acceleration).
3. From 60 to 70 seconds, the car decelerates uniformly; draw a line below the time axis with a constant negative acceleration until it comes to a complete stop.

The resulting acceleration-time graph distinctly shows sections of positive acceleration, zero acceleration, and negative acceleration.

To find the value of v, we consider the total distance:

• The distance covered during acceleration (first 20 seconds) can be calculated using the formula: $d_1 = \frac{1}{2} a t^2$
• During constant speed (60 seconds), the distance is: $d_2 = v \cdot 40$
• During deceleration (last 10 seconds), using similar formulas: $d_3 = \frac{1}{2} \cdot a_{deceleration} \cdot (10)^2$

We know: Total distance = 880 m, Thus, $d_1 + d_2 + d_3 = 880$ If $(d_1 + d_3) = \frac{1}{2} v \cdot 10$ (assuming the same magnitude of acceleration during both phases), then substituting gives:

$70 + 40 = \frac{v + 0}{2} \cdot 70$ Using the total distance relationship: $880 = 70 + v \cdot 40$ Solving yields:

• The value of v is 16 m/s.