Straight line motion question (1 Viewer)

[physicss]

Three cars A, B and C are traveling in the same direction on a straight section of a motorway, one in each lane. Car A moves with a constant velocity of 18m/s, while car B travels with constant acceleration of 0.5m/s^2 and car C has a constant acceleration of 1m/s^2. When first observed, car B is 100 metres behind car A and is traveling at 18m/s and car C is 162 metres behind car A and also traveling at 18m/s.

i) Show that at any time t seconds, car B is [100-((t^2)/4)] metres behind car A.

ii) Show that car C overtakes car A two seconds before car B overtakes car A.

iii) How far ahead of car A is car C at the moment when car B overtakes car A and what is the velocity of car C at that instant.

Need help/step by step solutions with all three please.

ANSWER for part iii) 38m ahead, 38m/s

 

[vincent3g]

Alright, this is like my first post in answering maths, so I can't do all the symbols.

So a diagram of the scene initially is
C--62--B---100---A
And you are given
Acceleration of C is
A{C}=1
and {x}-means subscript ;D
A{B}=0.5
so dV{B}/dt=0.5
Integrating both sides;
V{B}=0.5t + C
When t=0 V{B}=18 (given)
∴ 18=0.5*0 + C
C=18
Let the origin be at B's initial spot
so integrating that;
d{B}=0.5*(t^2/2)+18t+C
When t=0,d{B}=0, ∴ C=0
So the equation for the distance from the B's initial place has the formula
d=0.5*(t^2/2)+18t=t^2/4 + 18t
the Distance traveled by A is given by
d=100+18t (you should be able to see that, since v=18 and it started 100m from B)
So the distance between B and A, is d{A}-d{B}
=100+18t - t^2/4 -18t
=100-t^2/4 QED
ii)a{C}=1
dv{C}/dt=1
v{C} =t+C
When t=0 v{C}=18 (given)
∴v{C}=t+18
x{C}=t^2/2 + 18t+C
When t=0, x=-62 (taking initial B as the origin)
so; x{C}=t^2/2 + 18t-62
Now you have to find when C and B overtake A
i.e. Find when d{A}=d{C} and d{A} = d{B}
d{A}=d{C}
100+18t=t^2/2 + 18t-62
t^2 /2=162
t^2=324
t=18 (taking t>0)
and;d{A} = d{B}
100+18t=t^2/4 + 18t
t^2/4=100
t=20
and the differences between them is 2, QED
iii)Aim, find the d{C}-d{A},
when d{A}=d{B}
From above, it happens at 20s
so sub t=20 in
d{C}-d{A}=t^2/2 + 18t-62-(100+18t)
=38m
and sub in t=20 in v{C}
v{C}=t+18
v{C}=20+18=38m/s