Question

a driver takes 0.20 seconds to apply the brakes .if he is driving car at a speed of 54 km/hour and the brakes cases a deceleration of 6.0 ms find the distance travelled by the car after he see the need to put the brakes? ​

Answer 1

Answer:

5.35 metres

Explanation:

Given :

• Time taken = t = 0.2 seconds

• Initial velocity = u = 54 km/h = 54×5/18 = 15 m/s

• Final velocity = v = 0 m/s

• Acceleration = a = - 6.0 m/s²

To find :

• Distance travelled by the car after the brakes are applied

Using the third equation of motion :

V²-u²=2as

0²-15²=2×-6×s

-225=-12s

S=5.35 metres

The distance travelled by the driver is equal to 5. 35 metres with a Initial velocity 54 km/hr and final velocity is 0 m/s

More :

• First equation of motion = v=u+at
• Second equation of motion = s=ut+½at²

Answer 2

GIVEN :-

• Time (t) = 0.2 sec.

• Final velocity (v) = 0m/s .

• Initial velocity (u) = 54km/hr = 54 × 5/18 = 15m/s.

• Acceleration (a) = -6.0m/s² .

TO FIND :-

• The distance travelled by the car after the brakes are applied.

SOLUTION :-

Applying the 3nd equation of motion,

$\tt \: \dagger { \boxed {\red{ \tt \: v {}^{2} \: = \: u {}^{2} \: + \: 2as }}}$

[ Put the values ]

=> 0² = 15² + 2 × -6 × s

=> -225 = -12s

=> s = 5.35 m

Thus, the distance travelled by the car is 5.35 m after the brakes are applied.