A Motorcyclist travels on a straight road with a constant acceleration of 0.6 ms^-2 P and Q are two points on the road 120m apart. Given that the motorcyclist increases speed by 6 ms^-1 between these two points as she travels from P to Q find: a) The time taken to travel from P to Q
I think S=ut + 1/2at^2 S=120 , U=6 , A=0.6 , t= ? Is that correct?
You dont know u, as youre told the velocity increases by 6m/s over that time, so it cant be that one. Think about which suvats involve acceleration and time and whether you know the other "two" things.
You dont know u, as youre told the velocity increases by 6m/s over that time, so it cant be that one. Think about which suvats involve acceleration and time and whether you know the other "two" things.
S=vt - 1/2at^2. Or v=u + at that’s if u is 0. So which is it and what are the values for suvat?
Is it like this 6=0+0.6*t to get 10s ? What is ur answer for question?
Yes, though its incorrect to assume u=0. You should really just write it down as the definition of (constant) acceleration is acceleration = change in velocity / change in time so 0.6 = 6/t The v=u+at is the basic gcse suvat which wraps that up.
Yes, though its incorrect to assume u=0. You should really just write it down as the definition of (constant) acceleration is acceleration = change in velocity / change in time so 0.6 = 6/t The v=u+at is the basic gcse suvat which wraps that up.
Wow thanks, idk why it’s a 5 mark question just for that.