Mechanics 1 (2 questions form october november 2009) (Paper

[XWTBSILT]

Okay i want to know how to do q6
part 1 is ok i got the answer as 4m/s^2

in part ii i got 0.32m but the answer is 0.448m

can some one explain part ii and iii

and the other question is the last question
here it is

A motorcyclist starts from rest at A and travels in a straight line. For the first part of the motion, the
motorcyclist’s displacement x metres from A after t seconds is given by x = 0.6t2 − 0.004t3.

(i) Show that the motorcyclist’s acceleration is zero when t = 50 and find the speed V ms−1 at this time. [5]

For t ≥ 50, the motorcyclist travels at constant speed V ms−1.

(ii) Find the value of t for which the motorcyclist’s average speed is 27.5ms−1. [5] >>>>>>>>>>> i want to know how to do this only

i want to know how to do part ii only forst part is ok

Thanx in advance for ur replies

 

[PlanetMaster]

Re: Mechanics 1 (2 questions form october november 2009) (Pa

Oct/Nov 09 Paper 42 Question 6:
Part ii:
v² = u² + 2as
s = 1.6²/8
s = 0.32m (That's what you got)

Now question says B does not rebound upwards but that doesn't applies to A.
For the bump in A, v will be 0 at max height. u will be 1.6 and a will be g
v² = u² + 2as
0 = 1.6² - 2gs
s = 0.128m

So max height = 0.32 + 0.128
So max height = 0.448m

Part iii:
a = v/t
So t = v/a
So t = 1.6/4
So t = 0.4s

For that 0.128m covered when A bumped,
v = u + at
0 = 1.6 - 10t
t = 0.16s

Total time = 0.4 + 0.16
Total time = 0.56s

 

[Sharky]

Re: Mechanics 1 (2 questions form october november 2009) (Pa

ii) the answer is 0.448m this is because when particle B reaches the ground particle A will still go up due to the momentum that it is carrying so the solution will be,
s1: v=0 a=4 u=1.6
we use,
vsq=usq + 2as
0=1.6x1.6 + 2(4)s
s=0.32
s2: the particle will move the gravity
v=0 u=1.6
0=1.6x1.6 + 2(-10)s
s=0.128

Total height reached= 0.128+0.328
= 0.448m
iii) in this
we ttake time out for two parts of the journey,
for t1: v=u +at
1.6= 0 + 4t
t= 0.4s

for t2: v=u +at
0= 1.6 -10t
t=0.16s

Time: 0.16 + 0.4
0.56s

 

[XWTBSILT]

Re: Mechanics 1 (2 questions form october november 2009) (Pa

Thanx Guyz

 

[MAVtKnmJ]

Re: Mechanics 1 (2 questions form october november 2009) (Pa

Can someone do the motorcyclist one as well. Thanks!

 

[PlanetMaster]

Re: Mechanics 1 (2 questions form october november 2009) (Pa

Can someone do the motorcyclist one as well. Thanks!

Oct/Nov 09 Paper 42 Question 7:
Part ii:
As question says 'For t ≥ 50, the motorcyclist travels at constant speed V ms−1.' so the duration of acceleration was 50s.
Now we need time required to attain average speed of 27.5.

At 50s, distance covered by motorcyclist is 1000m (using equation from question).
Since average speed = total distance/total time’ and s = vt and v = 30 (from part i)
so 27.5 = (1000 + 30t)/(50 + t)
and therefore t = 150

Total time = 150 + 50
Total time = 200s

 

[MAVtKnmJ]

Re: Mechanics 1 (2 questions form october november 2009) (Pa

Thanks, I'll give it a try myself now!