PLS ANSWER VERY FAST URGENT !!!!!!!!!!!!!!!!!

Question

A newspaper seller is riding a bicycle with a pile of newspaper in his basket.

He drops the newspaper in the entrance of every house as he passes by

without slowing down his cycle. He passes house 203 with a velocity 42m/s

and constant acceleration and reaches house 204 in 16 sec. From house 204,

he continues travelling with triple the acceleration and reaches house 205

at a velocity 63m/s in 4 sec. What is the velocity of the newspaper seller at

house 204?

Answer 1

Hi "P":

Here's how to do this type of prob. You just need to know the laws of Newtonian motion.

Let seller's constant acceleration = a

Then v₂₀₃ = 42 m/s

So, v₂₀₄ = v₂₀₃ + at with t = 16 sec and v₂₀₃ = 42 m/s

=> v₂₀₄ = 42 + 16a . . . . . . . . . . . . Eqn (1)

Similarly, v₂₀₅ = v₂₀₄ + 3at since acceleration is now 3a

=> 63 = (42 + 16a) + 3a(4) from Eqn (1)

∴ 63 = 42 + 28a

21 = 28a

a = 21/28 = 3/4 m/s

And we're done!