Question

Example 8.7 The brakes applied to a car
produce an acceleration of 6 m sin
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

Given:-

• Acceleration ,a = -6m/s²

• Time ,t = 2s

• Final Velocity ,v = 0m/s

To Find :-

• Distance travelled by car, s

Solution:-

Firstly we calculate the initial velocity of the car

By using 1st equation of motion

• v = u+at

Substitute the value we get

→ 0 = u + -6×2

→ -u = -12

→ u = 12m/s

NOW USING 3RD EQUATION OF MOTION.

• v² = u² +2as

Substitute the value we get

→ 0² = 12² + 2× (-6) ×s

→ 0 = 144 + (-12) ×s

→ -144 = -12×s

→ s = -144/-12

→ s = 144/12

→ s = 12 m

Therefore , the distance covered by the car is 12 metres.

Answer 2

$\huge{\sf{\underline{\red{Given:}} }}</p><p>$

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 }

✒To Find :

• Distance (s) travelled by car.

$\rm\underline{\huge{\red{Solution:}}} </p><p>$

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

$\implies\:\sf{Initial\:velocity\:(u)\:=\:u}</p><p>$

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

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

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

$\implies\:\sf{\cancel{-12}s\:=\:\cancel{-144}} </p><p>$

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

• Therefore, the distance travelled by car is 12 m.