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

For the acceleration-time graph:

1. Mark the y-axis as acceleration (a in m/s²) and the x-axis as time (t in seconds).
2. For the first 20 seconds, the acceleration is positive and constant, marked as a positive uniform line.
3. From 20 seconds to 60 seconds, the acceleration is 0, represented by a horizontal line at 0.
4. From 60 seconds to 70 seconds, the acceleration is negative (deceleration) until the car comes to rest.

Key points to include:

• A constant positive acceleration for the first 20 seconds.
• Zero acceleration for the next 40 seconds.
• A constant negative acceleration during the last 10 seconds.

To find the value of v, we first determine the total distance moved by the car:

1. The distance during the acceleration phase (first 20 seconds) can be calculated as: d_1 = rac{1}{2} a t^2 Assuming constant acceleration, we can denote the final speed v after 20 seconds where: a = rac{v}{20} Therefore, d_1 = rac{1}{2} imes rac{v}{20} imes (20)^2 = 10v

2. The distance at constant speed (from 20 to 60 seconds) is: $d_2 = v imes 40$

3. During deceleration (from 60 to 70 seconds), the distance is: d_3 = rac{1}{2} v t = rac{1}{2} v imes 10 = 5v

Total distance is therefore: $d_{ ext{total}} = d_1 + d_2 + d_3 = 10v + 40v + 5v = 55v$

Given total distance is 880 m: $55v = 880$ Solving for v gives: v = rac{880}{55} = 16 \, m/s