Question

Example 8.7 The brakes applied to a car
produce an acceleration of 6 m s-2 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.finally can you explain in little bit breif​

Answer 1

Given :-

The brakes applied to a car produce an acceleration of 6 m/s⁻² in the opposite direction to the motion.

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

To Find :-

The  distance it travels during this time.

Analysis :-

Consider the initial velocity as a variable and substitute the values in the first equation of motion.

After finding the value of the initial velocity, substitute the values in the second equation of motion and find it accordingly.

Solution :-

We know that,

• a = Acceleration
• v = Final velocity
• u = Initial velocity
• t = Time

By the first equation of motion,

$\underline{\boxed{\sf v=u+at}}$

Given that,

Final velocity (v) = 0 m/s

Initial velocity = u

Since the acceleration shows negative sign, it's retardation.

Retardation = −6 m/s²

Time (t) = 2 sec

Substituting their values,

$\sf 0=u+(-6)(2)$

$\sf -u=-12$

$\sf u=12$

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

By the second equation of motion,

$\underline{\boxed{\sf s=ut+\dfrac{1}{2}at^{2} }}$

Given that,

Initial velocity (u) = 12 m/s

Time (t) = 2 sec

Retardation = -6 m/s²

Substituting their values,

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

$\sf s=24+(-3)(4)$

$\sf s=24-12$

$\sf s=12$

Therefore, the distance travelled by car is 12 m.