Question

A bus is travelling at speed of 15m/s. Then it accelerates for 5s at a rate of 10m/s2. Find the final velocity and distance covered by the bus during this process.

Answer 1

Given :-

Initial velocity = u = 15 ms-¹

Acceleration = a = 10 ms-²

Time taken = t = 5s

Now,

Using first equation of motion we can find the final velocity,

v = u + at

v = 15 + 10 × 5

v = 65 ms-¹

Again,

we can solve the distance covered by using both second and third equation of motion.

So using second equation,

S = ut + 1/2 at²

S = 15 × 5 + 1/2 × 10 × 5 × 5

S = 75 + 125

S = 200 m

Using third equation,

v² - u² = 2aS

(v-u)(v+u) = 2(10)S [v² - u² = (v-u)(v+u)

(65-15)(65+15) = 20S

20S = 4000

S = 200 m

Hence,

The final velocity = v = 65 ms-¹

Distance covered = S = 200 m

Answer 2

$\frak{Given}\:\begin{cases}\:\quad \sf Initial \:Velocity \:of\:bus\:,\:u\:=\:\pmb{\frak{ 15m/s}}\\ \:\quad \sf Acceleration \:of\:bus\:,\:a\:=\:\pmb{\frak{ 10\:m/s^2}}\\ \:\quad \sf Time\:Taken \:by\:bus\:,t\:=\:\pmb{\frak{ 5\:sec}}\:\end{cases}\:$

Need To Find : (I) Final Velocity and , (II) Distance Covered by bus ?

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

We've provided with the initial velocity (u) , acceleration (a) and time taken (t) by the bus , and we'll find final velocity (v) using First Equation of Motion that is : $\underline {\pmb{\sf \: v \:=\:u \:+\:at\:}}$

$\\ :\implies \sf v = u + at \:\\\\\\ :\implies \sf v = 15 + (\:10\:\times 5\:) \:\\\\\\:\implies \sf v = 15 + 50 \:\\\\\\ :\implies \sf v = 65 \:\\\\\\ :\implies \pmb {\underline {\boxed {\purple {\:\frak{ \:v\:\:=\:65\:m/s\:}}}}}\:\bigstar \:$

$\therefore \:\underline {\sf \: Hence,\:Final\:Velocity \:of\:bus\:is\:\pmb{\sf 65\:m/s\:}\:.}\:$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━━⠀

$\qquad \bigstar \:\underline {\sf According \:To\:The\:given \:Question \:\::\:}\:$

Now , We'll find Distance Covered by bus using Third Equation of Motion and that is : $\underline {\pmb{\sf \: v^2 \:- \:u^2\: \:=\:2as\:}}$

$\\\dashrightarrow \sf v^2 - u^2\:=\:2as\:\\\\\\ \dashrightarrow \sf (65)^2 - (15)^2\:=\:2\times \:(\: 10\:\times s\:)\:\\\\\\ \dashrightarrow \sf 4225 - 225\:=\:20s\:\\\\\\ \dashrightarrow \sf 4000\:=\:20s\:\\\\\\\dashrightarrow \sf 20s\:=\:4000\:\\\\\\\dashrightarrow \sf s\:=\:\cancel{\dfrac{4000}{20}}\:\\\\\\ \dashrightarrow \pmb {\underline {\boxed {\purple {\:\frak{ \:s\:\:=\:200\:m\:}}}}}\:\bigstar \:$

$\therefore \:\underline {\sf \: Hence,\:Distance \:Travelled \:by\:bus\:is\:\pmb{\sf 200\:m\:}\:.}\:$