Question

the brakes applied to a car produce an accelaratino of 6m/s2.in the opposition direction to the motion if the car take 2 second to stop after application of brakes calculate the distance​

Answer 1

Answer:

12 m

Explanation:

According to the given question,

-a = deceleration = 6 m/s^2

t = time = 2 sec

v = final velocity = 0 m/s

(As it stops finally)

s = displacement = ?

And we are asked to find the distance.

Here, one thing to note that distance is equal to the displacement, as the this motion is a linear motion.

we know,

• v = u+at

• 0 = u-6×2

• u = 12 m/s

And, again

• s = ut+1/2 × at^2

• s = 12×2 - 1/2 ×6 × 4

• s = 24-12

• s = 12 m ans.

Alternative

We have 5th formula of motion which is not written in many books,

• s = [(u+v)t]/2

• s = [(0+12)×2]/2

• s = 12×2/2

• s = 12 m ans.

Thus the car will take 12 m before stopping with the deceleration of 6 m/s^2 and time 2 sec.

Answer 2

ATQ,

acceleration = a = -6m/s^2 (∵ the acceleration is in the opposite of the direction of motion)

time taken to stop = t = 2 seconds

final velocity = v = o

initial velocity = u = ?

to find the distance, first we have to calculate it's initial velocity 'u'

by using 1st law of equation,

⇒ v = u + at

⇒ 0 = u + (-6 * 2)

⇒ -u = -12m/s

⇒ u = 12m/s

now, using 3rd law of motion,

⇒ s = ut + 1/2 at^2

⇒ s = (12 * 2) + 1/2 * -6 (2)^2

⇒ s = 24 - 3 (4)

⇒ s = 24 - 12

⇒ s = 12m

hence, the distance covered in 2 seconds by the cars is 12m.