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M1 doubts

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immortal:
Give me 10min to ans dem.

immortal:
These r answered 2 da best of my knowledge,so get dem re-checked specially Q.2 part c. & d.

Q1> solve wit eqn s=ut+0.5at2

as wen dey meet, v shud find da time for which dey cover da same distance

3t+0.5*4*t2-20=0t+0.5*3*t2
0.5t2+32-20=0
t=4(+ve value only)

s(x)=0.5*3*42
s(x)=24m

Q2>a.
s=122.5     u=0    a=9.8   t=?
s=ut+0.5at2
122.5=0t+0.5*9.8*t2
t=5  sec

b.
s=122.5   u=0   a=9.8     v=?
v2=u2+2*a*s
v2=0+2*9.8*122.5
v=49m/s

c.(not sure)
min value of "u" wen objects meet at surface.
dre4 s=122.5   t=5+2(assuming u wrote two)    a=9.8     u=?
s=ut+0.5at2
122.5=7u+0.5*9.8*72
u=12.6m/s

d.(not sure)
the Min value will depend of other factors, as time taken 4 da object 2 reach da surface will decrease.

Q3>a. & b.
    3g-T=3a
    T-2g=2a
if u substitute dem u get a=g/5 m/s(1.96)   & T=23.52

c.
a=1.96m/s2     u=0      s=1.5    v=?
v2=u2+2as
v2=0+2*1.96*1.5
v2=5.88

d.
dats da max distance moved by Q wit its existing force.
u=(speed with which p hits the ground)      a=-9.8     v=0       s=?
v2=u2+2*a*s
0=5.88+2*-9.8*s
s=0.3m(moved by q)
There4
2-0.3=1.7m closer 2 pully.

Phosu:
srry for the typo, it is 2 sec
but why did u add the secs...?

immortal:
sry 4 da mistake,
It shud b t=5-2
               =3
because as da stone reached da water surface at 5sec, da ball shut reach it at 3 sec, as its 2sec late.

halosh92:
could someone plz explain this???

a car of mass 1000kg exerts a driving force of 2.2kN when pulling a caravan of mass 500kg along a horizontal road.
The car and caravan increase speed from rest to 4m/s while traveling 16m.
given that the resistances on the car and caravan are proptional to their masses, find these resistances and the tension in the tow bar.

could someone specifically explain what they mean by the red colour sentence.

thx ASAP plzzzz

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