Question

A bus was moving with a speed of 54 km per hour. on applying brakes it stopped in 8 seconds. calculate the acceleration​

Answer 1

Step-by-step explanation:

I would do it this way:

We know out initial velocity (u) to be 54km/hr.

This comes to 15 m/s. (54 × 1000/3600) or

(54 ×5/18)

Thus u= 15 m/s.

Time taken to stop = 8 seconds.

Thus t = 8s.

The bus stops completely. Thus final velocity,(v) = 0 m/s.

Now, we can use the equation:

v=u+at

Now v = 0 , u = 15 and t= 8.

=>0=15+8a.

Rearranging,

=>a=−15/8.

Thus acceleration comes to −1.875m/s2.

Now we know acceleration, time and initial velocity. Thus distance can be calculated as:

s=ut+(at2)/2.

=>s=15×8+(−1.875×64)/2.

Solving this, we get the distance to be 60m.

P.S. The acceleration has a negative value because the value of initial velocity is greater than the value of final velocity. As the bus applies brakes.

This is known as deceleration.

Answer 2

Given: A bus was moving with a speed of 54 km per hour. on applying brakes it stopped in 8 seconds. calculate the acceleration.

$\Large {\underline{\gray{\frak { Answer \:\:: \:}}}}$

$\quad \dashrightarrow \sf{u=54k /hr }$

$\quad \dashrightarrow \sf{u=54 ×\dfrac{5}{18}}$

$\quad \dashrightarrow \sf u=\cancel{54} ×\dfrac{5}{\cancel{18}}$

$\quad \dashrightarrow \sf{u= 15m/s }$

$\quad \dashrightarrow \sf{v=0 }$

$\quad \dashrightarrow \sf{t= 8 sec}$

$\quad \dashrightarrow \sf{acceleration= \dfrac{v-u}{t} }$

$\quad \dashrightarrow \sf{a= \dfrac{0-15}{8} }$

$\quad \dashrightarrow \sf{a= \dfrac{-15}{8} }$

$\quad \dashrightarrow \sf{a= 1.875m/s^2}$

$\pmb{2\: equations\: of\: motion}$

$\quad \dashrightarrow \sf{s= ut + \dfrac{1}{2} at^2 }$

$\quad \dashrightarrow \sf{s= 15(8)+ \dfrac{1}{2} 1.875 (8)(8) }$

$\quad \dashrightarrow \sf{s= 120 + ( -60) }$

$\quad \dashrightarrow \sf{s= 120 -60) }$

$\quad \dashrightarrow \sf{s= 60 m}$

☆Hence, the answer is 60 meters☆