# Oxford PAT 2009, Question 27

## 7 thoughts on “Oxford PAT 2009, Question 27”

1. William Paisley says:

For part C, I assumed during that distance travelled he carried on starting and stopping, hence I used s=0.5t(U+V) where U is zero. Can you briefly explain why we assume constant velocity, thanks!

1. He does carry on starting and stopping, but the question says he rapidly accelerates to velocity v. This basically means that he gets to v in negligible time and so he is travelling at v for the time during which he is moving.

2. Saiyed says:

For part (f), why not include the energy needed for the car to travel 1km in addition to the energy transferred to the air (i.e. why not add your answer of 50kJ to 500kJ from part (d) ) ?

3. Hello, I’m quite unsure about the e part “calculate power required to overcome air resistance”. Don’t you also have to take into account the power needed to accelerate the car?

1. No you don’t have to. Have a look at the wording of the question. It says: ‘ … calculate the kinetic energy transferred to the air …’.

4. ZhouHuaiji says:

Hello, I’m quite confused about question(c), should total energy include energy to restart the car ? In that case , answer will be doubled: mv^2d/s

1. Can you explain your reasoning a bit more? I’m not sure I understand where your confusion is at the moment.