Question

A bus starting from rest moves with a uniform acceleration of 0.1 m s-2 for 2 minutes. Find (a) the speed acquired, (b) the distance travelled.

Answer 1

Initial velocity(u)="0m/s"

Accelaration(a)="0,1m/s-2"

Time="2 mins=60*2=120 secs"

Final velocity=?

v=u+at

v=0+(0.1)120

v=0+12

v=12m/s

i.e  

Final Velocity="12m/s"

Distance(s)=v(v)-u(u)=2a

s=144-0=2(0.1)

s=144=0.2

s=1440=2

s=720m

 

Therefore,

(a)=12m/s

(b)=720m

Answer 2

Hello

Answer

Given ,

Here initial velocity ( u ) = $0ms^-1$

Acceleration ( a ) = $0.1ms^-2$ =

Time (t) = 2 minutes = 2 * 60 = 120 s

(a)  From the equation of motion , v = u + at = $0ms^-1$ +$0.1ms^-2$ * 120s = $12ms^{-1}$

(b) we Know that

$v^2$ -  $u^2$ = 2as

Here u = 0 ;

$v^2$ -  $0$ = 2as

⇒ $s = v^2/2a$ = $(12 ms^-{2} )/ 2*0.1 ms^{-2}$

⇒ $144m^2s^{-2} /0.2ms^{-2}$

⇒ 720 m

The speed acquired by the bus is 12ms^ -1 and the distance travelled is 720 m

Aliter : We know

s = ut+ 1/2 at^2

⇒ $0ms^{-1} *120s + 1/2 *0.1ms^{-1} * (120 s ) ^2$

⇒ 1/2 * 0.1 * 120 * 120

⇒ 720 m