Question

a) derive the equation for velocity - time relation graphically

b) a train starting from rest attains a velocity of 72km/h in 5 min. Assuming that the acceleration is uniform,find(i) the acceleration and the (ii) the distance travelled by the train for attaining this velocity




please explain this both questions very clearly I am having exam tomorrow please................
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Answer 1

In the Above attachments I have Derived 2 equation of Motion.

Deriving third equation of motion i.e ( v² - u² = 2as).

To Find this, See Fig:- 1

Area of Trapezium = OABC

Distance\:Travelled = Sum of Parallel Sides × Distance Between them = ½ × BC + OA × AD

So,

$\implies\rm{\dfrac{1}{2}\times{v + u}\times{t}}$

As, We know that In third equation of motion there is no "t".

eliminating "t" by using First Equation motion,

i.e, V = u + at

or a = $\rm{ t = \dfrac{v - u}{a}}$

Now, Putting the value of "t".

$\implies\rm{\dfrac{1}{2}\times{v + u}\times{t}}$

$\implies\rm{S=\dfrac{1}{2}\times{v + u}\times{\dfrac{v -u}{a}}}}$

$\implies\rm{2s=\dfrac{v^{2} - u^{2}}{a}}$

$\implies\rm{ 2as = v^{2} - u^{2}}$

Hence, We Derived the Third Equation of motion easily.

B).

Given:-

• Initial Velocity = 0m/s (As it was in rest)

• Final Velocity = 72km/h = 20m/s

• Time = 5min = 300s

To Find:-

• The Acceleration and the Distance Travelled.

FORMULAE USED

• v = u + at, or a = v - u/t

• S = ut + ½ × a × t²

v = Final Velocity

u = Initial Velocity

a = Acceleration

t = Time

s = Distance

Now,

v = u + at:, or

a = v - u/t

a = 20 - 0/300

a = 20/300

a = 0.06m/s²

Hence, The Acceleration is 0.06m/s²

Therefore,

S = ut + ½ × a × t²

S = (0)(300) + ½ × 0.06 × 90000

S = 0 + 0.03 × 90000

S = 2700m

Hence, The Distance travelled by train is 2.700m or 2.7km