Question

2. The brakes applied to a car produce an acceleration of 6 m /s2 in the opposite direction to the motion. If the car takes 2 s to stop after the application of brakes, calculate the distance it travels during this time

Answer 1

Answer:

here is your answer

Explanation:

Acceleration a = -6 m/s^2

Time t = 2 s

Final velocity v = 0 m/s

Let initial velocity be u

Let distance be s

v = u + at

So, 0 = u + (-6)(2)

So, u = 12 m/s

Now, s = ut + (1/2) at^2

So, s = 12(2) + (1/2)(-6)(2^2)

So, s = 24 - 12

So, s = 12 m

Thus, distance travelled is 12 m

hope u understand

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Answer 2

$\underline{\red{Given:}}$

The brakes applied to a car produce an acceleration of 6 m/s² in the opposite direction to the motion.

{ Acceleration i.e. a is -6 m/s²

Negative sign shows retardation }

The car takes 2 s to stop after the application of brakes.

{ Time i.e. t is 2 sec and Final velocity i.e. v is 0 m/s }

$\underline{\red{To\:Find:}}$

• Distance (s) travelled by car.

$\underline{\huge{\red{Solution:}}}$

Let us assume that the car is travelling with an initial velocity of u.

$\implies\:\sf{Initial\:velocity\:(u)\:=\:u}$

Using the First Equation Of Motion,

$\implies\:\sf{v \:=\: u + at}$

Substitute the known values

$\implies\:\sf{0\:=\:u+(-6)(2)}$

$\implies\:\sf{\cancel{-}u\:=\:\cancel{-}12}$

$\implies\:\sf{u\:=\:12}$

Therefore, the initial velocity of the car is 12 m/s.

Now, using the SECOND EQUATION OF MOTION,

$\implies\:\sf{s\:=\:ut+1/2at^2}$

Substitute the known values

$\implies\:\sf{s\:=\:12(2)+1/2(-6)(2)^2}$

$\implies\:\sf{s\:=\:24+(-3)(4)}$

$\implies\:\sf{s\:=\:24-12}$

$\implies\:\sf{s\:=\:12}$

OR

Using the THIRD EQUATION OF MOTION,

$\implies\:\sf{v^2-u^2\:=\:2as}$

Substitute the known values

$\implies\:\sf{(0)^2-(12)^2\:=\:2(-6)(s)}$

$\implies\:\sf{0-144\:=\:-12s}$

$\implies\:\sf{\cancel{-12}s\:=\:\cancel{-144}}$

$\implies\:\sf{s\:=\:12}$

Therefore, the distance travelled by car is 12 m.