Question

A cyclist is in a race and 100m from the finish line and decides to accelerate his speed. he maintains a constant acceleration of 0.32m/s^2. If a crosses the finish line with the speed of 17m/s. how fast was he cycling when he started to accelerate.

Answer 1

▶ Solution :

Given :

▪ Distance travelled = 100m

▪ Acceleration of cyclist = 0.32m/s^2

▪ Final velocity of cyclist = 17mps

To Find :

▪ Initial velocity of cyclist.

Concept :

✒ Since, acceleration is constant throughout the whole journey, we can apply equation of kinematics directly.

✒ Third equation of kinematics is given by

☞ v^2 - u^2 = 2as

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• s denotes distance

Calculation :

→ v^2 - u^2 = 2as

→ (17)^2 - u^2 = 2(0.32)(100)

→ 289 - u^2 = 64

→ u^2 = 289 - 64

→ u^2 = 225

→ u = √(225)

→ u = 15mps

Answer 2

Given :-

Distance covered by cyclist(s) = 100m

Acceleration (a)= 0.32m/s^2

Final velocity (v)=17m/s

To find :-

The initial speed (u) of cyclist

$\underline{\bigstar{\sf{\ By\ Third\ equation\ of \ motion}}}$

$\boxed{\sf{\dag\ \ v^2=u^2+2as}}$

● Where !

s = Distance

u= Initial velocity

v= final velocity

a= acceleration

Now find the initial velocity of cyclist

$:\implies\sf v^2= u^2+2as\\ \\ :\implies\sf (17)^2= u^2+2\times 0.32\times 100\\ \\ :\implies\sf 289= u^2+ 0.64\times 100 \\ \\ :\implies\sf 289= u^2+64\\ \\ :\implies\sf 289-64= u^2\\ \\ :\implies\sf 225=u^2\\ \\ :\implies\sf \sqrt{225}=u\\ \\ :\implies\sf 15=u\\ \\ :\implies \sf u= 15m/s$

$\underline{\boxed{\textsf{\textbf{ Initial \ velocity \ of \ cyclist = 15m/s}}}}$