AQA Mechanics (M1) Help :(

Speed is distance over time
So 100/9.83 give 10.17m/s
I dont know how to do part b though.

Posted from TSR Mobile

Just use the constant acceleration equations
U=0
V=16.3
s=60
seriously havent done this for a year and i done edexcel but mechanics is universal.

what equation did you use?

V^2=u^2+2as
they need to be memorised, or at least in edexcel that had to be.

If the acceleration isn't given for the first 60m, how did you manage to calculate the speed?

I really wish i could help you but ive gotta get on with my revision ill write it down and do it in a break or something and if i get an answer ill post back here.

This is actually a far more complicated question that it might at first seem. The run consists of two parts: the first 60m with constant acceleration(a) and the last 40m at constant speed(v). The whole 100m takes 9.83 seconds. To solve this w)e need to split the time into the time for the first 60m(T, and the time for the last 40m(9.83-T). The sprinter has an initial speed of zero.

We have two unknowns we need to find, a and v, and the another unknown, which is the time for the first 60m(T) so we will need 3 simultaneous equations:

We know that the total distance travelled is a hundred and this equals:

100= 60(the first bit) + 40(the second bit)


1.the first bit= T(u+v)/2, and since u=0 ->60=Tv/2

2.the second bit = (9.83-T)v ->40=(9.83-T)v

3.and the acceleration is: a=v/T since u=0

Now, we need to rearrange the equations to eliminate T, which we don't want. part a) asks for v, and we can see that equations 1. and 2. contain T and V, but no a.

we can rearrange those and solve for v:

60=Tv/2 -> T=120/v

Sub in to 2.
40=(9.83-120/v)v -> 9.83v-120=40 -> v=160/9.83 ->v=16.2767... so 16.3 ms-1 to 3sf


Now we can solve for a,
a=v/T
and using 1. again, 60=Tv/2 -> T=120/v

Sub in:

a=v/(120/v) -> a=v^2/120 -> Plug in 16.3 -> a= 16.3^2/120 = 2.214... is 2.21 ms-2 to 3 sf

Could be a better explanation. Could be done with a speed-time graph, but idk if AQA do those. The key thing is splitting it into two bits with and realising that there's an unknown time for how long each bit lasts

THANK YOU EVER SO MUCH FOR YOUR HELP! IT IS VERY MUCH APPRECIATED!

However, i'm a bit confused by this one point:

"2.the second bit = (9.83-T)v ->40=(9.83-T)v "

have you assumed that u = 0, because the athlete is still running so their initial speed would be the speed calculated from the first 60m