Question

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Answer 1

(ii)

Answer 2

(a)
v1=v0+at
so
t=(v1-v0)/a=(6-3)/0.06=50s

Answer 3

i got the a and b part.
but im stuck with (ii)

Answer 4

Since I need the distance for (ii), I'll finish (b)
v1^2-v0^2=2aS
S=(v1^2-v0^2)/(2a) = (6^2-3^2)/(2*.06)= 225m

Answer 5

i got the (b) correct as well. :)

Answer 6

(ii) (a)
\[Distance = \int\limits vdt = \int\limits_0^{50} kt^2dt = \left[ kt^3/3 \right]_0^{50} = 225\]
k=225*3/50^3=0.0054

Answer 7

i can't get it. =(
is there any other way?

Answer 8

(b)
v1=k(50)^2=0.0054*50^2=13.5 m/s

Answer 9

is there any another way?

Answer 10

I am not sure what you mean by "not get it". Is it the method, or the calculations?

Answer 11

method..
is there any another method? where we dont use Calculus?

Answer 12

In kinematics, there are standard formulas that apply to uniform acceleration, like in part (i). For linear acceleration (v=kt^2, so a=2kt) I haven't seen any. If there is any, they are derived from calculus.
Are you not supposed to use calculus for that? If that's the case, I'll try to cook up something.

Answer 13

Alright. sure.
Thanks!! :)

Answer 14

heading to bed now.. c ya.. tc.

Answer 15

You're welcome, good night! :)

Answer 16

same 2 u. thanks.