8 (a) An electric car is travelling at a speed of 16.0 m/s - Edexcel - GCSE Physics - Question 8 - 2023 - Paper 2

To calculate the time taken (t) for the battery to become discharged, we use the formula:

$t = \frac{E}{P}$

Where:

• E = total energy transferred = 126 MJ = 126 \times 10^6 J
• P = average power = 17.5 kW = 17.5 \times 10^3 W

Substituting the values:

$t = \frac{126 \times 10^6}{17.5 \times 10^3}$

Calculating this gives:

$t ≈ 7200 \text{ seconds}$

To convert seconds to hours:

$t = \frac{7200}{3600} = 2 \text{ hours}$

The electrical device serves dual purposes: as a motor during acceleration and as a dynamo during deceleration.

When the car accelerates, the device uses energy from the battery, contributing to the car's movement. However, during deceleration, the device acts as a dynamo, converting kinetic energy back into electrical energy. This regenerated energy can be fed back into the battery, thus recharging it.

By harnessing energy that would otherwise be wasted during braking or coasting, the device effectively extends the car's driving range by increasing the energy available in the battery.

To comment on the claim, we first calculate the time using the provided formula:

$t = \frac{E}{I \times V}$

Where:

• E = 126 MJ = 126 \times 10^6 J
• I = current = 15.0 A
• V = voltage = 400 V

Substituting the values:

$t = \frac{126 \times 10^6}{15.0 \times 400}$

Calculating this gives:

$t = \frac{126 \times 10^6}{6000} = 21000 \text{ seconds}$

Converting seconds into hours:

$t = \frac{21000}{3600} \approx 5.83 \text{ hours}$

Since 5.83 hours is less than 6 hours, the claim is justified.

To calculate the total charge (Q) that moves into the battery, we use the formula:

$E = Q \times V$

Rearranging for Q gives:

$Q = \frac{E}{V}$

Substituting the values:

$Q = \frac{126 \times 10^6}{400}$

Calculating this gives:

$Q = 315000 \text{ C}$