Question

a driver takes 0.2s to apply the breaks. if he is driving car at a speed of 54kmh-1 and the breaks cause a acceleration of 6ms-2. find the distance?

Answer 1

Answer:

Explanation:

Solution:

Initial velocity=u=54km/h=54x5/18=15m/s

time=t=0.2sec

Final velcoity=v=0m/s

deceleration =a= 6m/s2

Now, Distance travelled by car during reaction time : 

Distance=Speedx time

Distance=15x0.2

=3m

From third equation of motion:

v²-u²=2as

0²-15x15=2x-6xs

s=-15x15/-12

S=225/12a

s=18.75m 

So the total distance travelled by the driver after  he seeds the need to put the brakes ON= 18.75+3=21.75m

OR

Solution:

Initial velocity=u=54km/h=54x5/18=15m/s

time=t=0.2sec

Final velcoity=v=0m/s

deceleration =a= 6m/s2

Now, Distance travelled by car during reaction time : 

Distance=Speedx time

Distance=15x0.2

=3m

From third equation of motion:

v²-u²=2as

0²-15x15=2x-6xs

s=-15x15/-12

S=225/12a

s=18.75m 

So the total distance travelled by the driver after  he seeds the need to put the brakes ON= 18.75+3=21.75m