Question

the brakes applied to a car produce an acceleration of 6m /s2 in the opposite direction of the motion if the car takes 2 second to stop after the application of brakes.calculate the distance travelled this time.

Answer 1

a=-6m/s^2
t=2sec
v=0m/s
u?
s?
v=u+at
u=v-at
u=0-(-6×2)
u=12m/s


s=v^2-u^2/2a
s= 0-144/-12
s=12m

Hope it helped mark as brainlist

Answer 2

$\huge\tt{Answer:-}$

$\bf{Given:-}$

• Negative Acceleration = $\tt{ 6m \ {s}^{-1} }$ $(a)$
• Time taken to stop = $\tt{ 2s}$ $(t)$

And,

• Final Velocity = $\tt{0m \ {s}^{-1}}$ $(v)$.

(Since it will slow down and will stop.)

____________...

Finding Initial velocity $(u)$ :-

We know,

Acceleration = (Final Velocity - Initial Velocity)/Time taken

Or,

$\tt{a = \frac{v-u}{t} }$

Putting the values:

$\tt{ \implies \frac{- 6m}{{s}^{2}} = \frac{0 - u}{2s}}$

$\tt{ \implies \frac{-u}{2s} = \frac{-6m}{{s}^{2}}}$

$\tt{ \implies -u = \frac{-6m}{{s}^{2}} \times 2s}$

$\tt{ \implies -u = \frac{-6m}{s \times \cancel{s}} \times 2 \cancel{s}}$

$\tt{ \implies -u = \frac{-6m}{s} \times 2}$

$\tt{\implies -u = -12 m {s}^{-1} }$

$\tt{\implies u = 12 m {s}^{-1} }$ (Initial Velocity)

______________...

Finding distance $(s)$ travelled :-

We know,

$\tt{ {v}^{2} - {u}^{2} = 2as }$

$\tt{\implies - {u}^{2} = 2as }$

$\tt{ \implies s = \frac{- {u}^{2}}{2a}}$

Substituting the values, we get:

$\tt{\implies s = \frac{- {12 m \ {s}^{-1}}^{2}}{2 \times \frac{- 6}{{s}^{2}}}}$

$\tt{\implies s = \frac{12 m \times 12 m}{2 \times 6m} (s) (s) (\frac{1}{s}) (\frac{1}{s})}$

$\boxed{\tt{\implies s = 12m }}$ $...(Ans.)$

#PhysicsBaba!