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# M1 Question watch

1. Hi,

Could someone please try this question and tell me what they get?

For part i) I get t=100

and part ii) I get 400m

Thanks

Hawk
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2. I get the same.
3. Taking A:
a(t)=0.4
v(t)=0.4t+c and as it starts from rest c=0 so v(t)=0.4t
x(t)=0.2t^2+d and as we are taking this place as the origin, d=0 so, for a:
X(T)=0.2t^2

For B:
a(t)=0.2
v(t)=0.2t+6
x(t)0.1t^2+6t+400

When they meet:
0.2t^2=0.1t^2+6t+400
0.1t^2-6t-400=0
t^2-60t-4000=0
t=(60+(3600+16000)^0.5)/2
=100s (as when you take away the square root, you get a negative value of t and t>0 for this)

When t=100, A is 0.2x100^2 from where it started (2km) and it started 2.4km from P so it is 0.4km from P when the lorries draw level.

So yes, I agree.
4. ty,

it was a question on todays M1 paper today btw (CCEA)
5. I was gonna say - thought all the M1 papers were over by now

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