Question

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 breaks calculate the distance it travel during this time.​

Answer 1

Answer:

Given:

➡️ acceleration= -6m/s^2

➡️time taken to stop= 2 s

➡️ it stops at t=2 s

so, at that time, it's velocity will be 0 m/s

To find :

➡️distance travelled during this 2 s

Solution :

We know that

v= u + at

so, here,

0= u-12

so, u= 12 m/s

also,

s = ut + 1/2 at^2

s= 12×2 + 1/2 ×-6 ×2×2

s=12m

hence, he travels 12 m in this 2 sec

Answer 2

Answer:

Given :-

• The brakes applied to a car produce an acceleration of 6 m/s² in the opposite direction to the motion.
• The car takes 2 seconds to stop after the application of brakes.

To Find :-

• What is the distance it travel during this time.

Solution :-

First, we have to find the initial velocity :

Given :

• Acceleration = - 6 m/s²
• Time Taken = 2 seconds
• Final Velocity = 0 m/s

According to the question by using the formula we get,

$\implies \bf v =\: u + at$

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

$\implies \sf 0 =\: u + (- 12)$

$\implies \sf 0 =\: u - 12$

$\implies \sf 0 + 12 =\: u$

$\implies \sf 12 =\: u$

$\implies \sf\bold{\blue{u =\: 12\: m/s}}$

Now, we have to find the distance travelled the car during this time :

Given :

• Initial Velocity = 12 m/s
• Final Velocity = 0 m/s
• Acceleration = - 6 m/s²

According to the question by using the formula we get,

$\leadsto \bf v^2 =\: u^2 + 2as$

$\leadsto \sf (0)^2 =\: (12)^2 + 2(- 6)s$

$\leadsto \sf 0 =\: 144 - 12s$

$\leadsto \sf 0 - 144 =\: - 12s$

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

$\leadsto \sf 144 =\: 12s$

$\leadsto \sf \dfrac{144}{12} =\: s$

$\leadsto \sf 12 =\: s$

$\leadsto \sf\bold{\red{s =\: 12\: m}}$

$\therefore$ The distance travelled during this time is 12 m .