Question

As a car passes the point A on s straight road, its speed is 10 m/s. The car moves with constant acceleration a m/s² along the road for T seconds until it reaches the point B, where its speed is V m/s. The car travels at this speed for a further 10 seconds, when it reaches point C. From C it travels for a further T seconds with constant acceleration 3a m/a² until it reaches a speed of 20 m/s at the point D. Show that V= 125. Given that the distance between A and D is 675 m, Find the values of a and T.

Answer 1

maybe start by drawing Velocity-time graph

Answer 2

x₁ = 0 + 10T + ½aT²

v² = v₀²+2as
vB² = 10²+ 2a(10T + ½aT²)
vB = √[10²+ 2a(10T + ½aT²)]

vC = VB = √[10²+ 2a(10T + ½aT²)] ◄─── Eq # 1

x₂ = x(B→C) = 10 * vB
x₂ = 10 * √[10²+ 2a(10T + ½aT²)]


v = v₀ + at
20 = vC + (3a)T
vC = 20 - 3aT ◄─── Eq # 2

Answer 3

solve to find T

Answer 4

iam not that good on drawing graphs

Answer 5

okay, its not required

Answer 6

first lets try solving for T, using the fact that,
time between A->B and C->D is same

Answer 7

okay

Answer 8

scratch that,

first lets try solving for \(\color{red}{V}\), using the fact that,
time between A->B and C->D is same

Answer 9

A->B :
V = 10 + aT ---------------(1)

C->D :
20 = V + (3a)T --------------(2)

Answer 10

solve them both for V

Answer 11

V=10 +aT
20=10+at+3at
10=4at
t=10/4a
t=5/2a

Answer 12

substitute that value of t in first equation,
that gives u value of V

Answer 13

V=10+a(5/2a)
V=10+5a/2a
V= 10+5/2
V=25

Answer 14

sorry 12.5

Answer 15

yes

Answer 16

Now we could find the distances am i right

Answer 17

oh wait, we need to find \(a\) and \(T\) still

Answer 18

oh i forgot

Answer 19

u familiar wid below equation :
v^2 - u^2 = 2*a*s ?

Answer 20

say, the distance between A->B is \(S1\),
the distance between C->D is \(S2\)

then,
since the total distance is 625,
\(S1 + 10*12.5 + S2 = 625\) ----------(3)

Answer 21

A->B :
\(12.5^2 - 10^2 = 2*a*S1\) -----------(4)

C->D :
\(12.5^2 - 10^2 = 2*(2a)*S2\) --------(5)

Answer 22

eliminate \(a\) from equations (4) and (5)

Answer 23

then, u solve S1 and S2 distances

Answer 24

had we graphed Velocity-time graph, this wud have been much simpler to calculate :o

Answer 25

what is a equal to

Answer 26

but ok,
(3), (4), (5) are not looking that scary, u can solve them easily :)

Answer 27

eliminate \(a\) by dividing :

(4) / (5)

Answer 28

when i looked back at the book a=1/8

Answer 29

how?

Answer 30

we wil wry about \(a\) later,
first solve the distances \(S1\) and \(S2\)

Answer 31

(5)/ (4) :

\(\large \frac{20^2-12.5^2}{12.5^2-10^2} = \frac{3S2}{S1}\)

\(13S1 = 9S2\)------------(6)

Answer 32

u can solve (3) and (6) for \(S1\) and \(S2\)
once u have distances, u can find \(a\) and \(T\)

Answer 33

http://www.wolframalpha.com/input/?i=S1+%2B+10*12.5+%2B++S2+%3D+675%2C+12.5%5E2+-+10%5E2+%3D+2*a*S1%2C+20%5E2+-+12.5%5E2+%3D+2%283a%29S2