Question

the brakes applied to a car produce an acceleration of 6m/s2 in opposite direction to the motion. if the car take 2 second to stop after the application of break, calculate the distance its travel during this time ?

Answer 1

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

Answer 2

Answer:

HERE , the acceleration is in opposite direction that means it's like a frictional force.

So, if acceleration is opposite then

a = -6m/s

where a is acceleration.

t = 2s

where t is time .

After applying the brakes the car is stopped......means final velocity (v) is 0.

v = 0m/s

where v is final velocity .

As we know, that

v = u + at

0 = u + (-6) × 2

0 = u - 12

u = 12

then ,

s = ut + at ^2

2

= 12 × 2 + (-6) × 2^2

2

= 24 - 6 × 4 ÷ 2

= 24 - 24 ÷ 2

= 24 - 12

= 12m/s

So , after the brake has been applied the car stops after moving 12 m.

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