ExpertHard6 marksExtended Response
AQA GCSE · Question 07.4 · Forces
Another train travels at a speed of 60 m/s.
A constant braking force of 270 000 N causes the train to decelerate and stop.
mass of train = 240 000 kg
Calculate the distance travelled while the braking force is applied.
Use the Physics Equations Sheet.
Another train travels at a speed of 60 m/s.
A constant braking force of 270 000 N causes the train to decelerate and stop.
mass of train = 240 000 kg
Calculate the distance travelled while the braking force is applied.
Use the Physics Equations Sheet.
How to approach this question
This is a multi-step problem. You can solve it in two main ways.
**Method 1 (Forces and Motion):**
1. Use Newton's Second Law (F=ma) to find the deceleration caused by the braking force.
2. Use the equation of motion that links initial velocity (u), final velocity (v), acceleration (a), and distance (s). (Hint: v² = u² + 2as).
3. Rearrange the equation to solve for distance (s). Remember that deceleration is a negative acceleration.
**Method 2 (Energy):**
1. Calculate the initial kinetic energy of the train (KE = ½mv²).
2. The work done by the braking force must be equal to the initial kinetic energy to bring the train to a stop.
3. Use the work done equation (Work = Force × distance) and set it equal to the kinetic energy.
4. Rearrange to solve for distance.
Full Answer
This problem can be solved by combining equations for force and motion, or by using the concept of work and energy.
**Method 1: Using F=ma and Equations of Motion**
1. **Calculate deceleration (a):**
Use Newton's Second Law, F = ma. Rearrange to find a:
a = F / m
a = 270 000 N / 240 000 kg
a = 1.125 m/s²
Since this is a braking force, the acceleration is negative: a = -1.125 m/s².
2. **Calculate distance (s):**
Use the equation of motion that relates initial speed (u), final speed (v), acceleration (a), and distance (s):
v² = u² + 2as
We know:
u = 60 m/s (initial speed)
v = 0 m/s (the train stops)
a = -1.125 m/s²
Rearrange to solve for s:
s = (v² - u²) / 2a
s = (0² - 60²) / (2 × -1.125)
s = (-3600) / (-2.25)
s = 1600 m
**Method 2: Using Work-Energy Theorem**
1. **Calculate initial Kinetic Energy (KE):**
The work done by the brakes must equal the initial kinetic energy of the train to bring it to a stop.
KE = ½mv²
KE = 0.5 × 240 000 kg × (60 m/s)²
KE = 120 000 × 3600
KE = 432,000,000 J
2. **Calculate distance (d):**
Work Done (W) = Force (F) × distance (d)
W = KE
Fd = 432,000,000 J
d = 432,000,000 J / 270 000 N
d = 1600 m
Common mistakes
✗ Mixing up initial (u) and final (v) velocities.\n✗ Forgetting to use a negative sign for deceleration in the motion equation.\n✗ Making calculation errors, especially with large numbers or squaring.\n✗ Using the wrong equations.
Practice the full AQA GCSE Physics Higher Tier Paper 2
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