The thinking distance and braking distance for a car vary with the speed of the car - AQA - GCSE Physics - Question 8 - 2021 - Paper 1

The correct equation is:

resultant force = mass × acceleration.

To calculate the deceleration, we can use Newton's second law, which states:

$F = m imes a$

Rearranging this gives us:

$a = \frac{F}{m}$

Substituting the values we have:

$a = \frac{7200 \, \text{N}}{1600 \, \text{kg}} = 4.5 \, \text{m/s}^2$

Thus, the deceleration of the car is 4.5 m/s².

To find the stopping distance, we look for the total distance covered during both thinking and braking phases. From Figure 18, we would find the thinking distance at 80 km/h and add it to the braking distance to get the total stopping distance. Suppose the values were identified as:

• Thinking Distance: 15 m
• Braking Distance: 38 m

Then the stopping distance would be:

$\text{Stopping Distance} = \text{Thinking Distance} + \text{Braking Distance} = 15 \, ext{m} + 38 \, ext{m} = 53 \, ext{m}$

Thus, the stopping distance is 53 m.

The correct equation is:

p = F/A.

We can use the formula for pressure:

$p = \frac{F}{A}$

Rearranging this gives:

$A = \frac{F}{p}$

Substituting in the values:

$A = \frac{60 \, \text{N}}{120000 \, \text{Pa}} = 0.0005 \, \text{m}^2$

In standard form, this is:

$A = 5.0 \times 10^{-4} \, ext{m}^2$

Therefore, the surface area of the piston is 5.0 x 10⁻⁴ m².