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

To find the time taken for the battery to become discharged, we use the equation:

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

Where:

• $E = 126 \, \text{MJ} = 126 \times 10^6 \, \text{J}$ (energy transferred)
• $P = 17.5 \, \text{kW} = 17.5 \times 10^3 \, \text{W}$ (power)

Substituting the values:

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

Calculating this gives:

$t = 7200 \, \text{s}$

Now converting seconds to hours:

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

Thus, the time taken for the battery to be discharged is 2 hours.

When the car decelerates, the electrical device acts as a dynamo, converting kinetic energy back into electrical energy, which can be stored in the battery. This process harnesses energy that would otherwise be wasted during braking.

By converting this kinetic energy to electrical energy, the battery is recharged, thus increasing its energy storage capacity. The more the battery is recharged during deceleration, the longer the car can be driven before the battery becomes fully discharged.

To evaluate the claim, we first calculate the time taken to transfer 126 MJ of energy:

Using the equation:

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

Where:

• $E = 126 \times 10^6 \, \text{J}$
• $I = 15.0 \, \text{A}$
• $V = 400 \, \text{V}$

Substituting in the values:

$t = \frac{126 \times 10^6}{15 \times 400}$ $t = \frac{126 \times 10^6}{6000}$ $t = 21000 \, \text{s}$

Converting seconds to hours:

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

Since 5.83 hours is less than 6 hours, the claim is justified, as the transfer can indeed occur in less than 6 hours.

To calculate the total charge transferred to the battery, we use the equation:

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

Where:

• $E = 126 \times 10^6 \, \text{J}$
• $V = 400 \, \text{V}$

Substituting these values:

$Q = \frac{126 \times 10^6}{400}$ $Q = 315000 \, \text{C}$

Thus, the total charge that moves into the battery while it is being recharged is 315,000 C.