Question

. a person is moving on a bike along a straight line ab. when he just passes a accelerates his bike uniformly at 1m/s² to reach b. he arrives at b with a speed of 108 kmph after 20s. the length ab is a) 300m b) 400m c) 500 m c) 700m​

Answer 1

Answer:

b) 400

Explanation:

You can use s=ut +1/2*at^2

Points to note

'u' will be zero because a part of a motion is considered

Convert 108 kmph to m/s

Answer 2

$\bigstar\:\:\:\rm\blue{GIVEN}\:\:\:\bigstar$

• $\rm\blue{Final\:Velocity=108km/hr=30m/s}$

• $\rm\red{Acceleration=1m/s^{2}}$

• $\rm\pink{Time\:Taken=20s}$

• He Moves in a Straight line from A to B.

$\bigstar\:\:\:\rm\blue{To\:Find}\:\:\:\bigstar$

• The length(Distance) AB covered by Bike.

$\bigstar\:\:\:\rm\green{FORMULAE\:USED}\:\:\:\bigstar$

• ${\boxed{\rm{\blue{v=u+at}}}}$

Where,

V= Final Velocity

u= Initial Velocity

t= Time

a= Acceleration

• ${\boxed{\rm{\blue{S=ut+\dfrac{1}{2}a(t)^2}}}}$

Where,

S= Distance

u= Initial Velocity

t= Time

a= Acceleration

Now,

• Using first equation of motion to find initial Velocity

$\implies\rm\green{v=u+at}$

$\implies\rm\green{30=u+1\times{20}}$

$\implies\rm\green{u=10m/s^1}$

Now,

$\implies\rm\blue{S=ut+\dfrac{1}{2}a(t)^2}$

$\implies\rm\blue{S=10×20+\dfrac{1}{\cancel{2}}\times{a}\times{\cancel{400^{200}}}}$

$\implies\rm\blue{S=200+200}$

$\implies\rm\blue{S=400m}$

Thus, The distance Covered by Person is 400m.