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A driver takes 0.20 s to apply the brakes after he sees need for it. If he is driving a car at speed of 54 km/h and the brakes cause deacc. of 6 m/s^2,find the distance travelled by car after he sees need to put brakes. (ANSWER GIVEN: 22 m) JUSTIFY your answer

A driver takes 0.20 s to apply the brakes after he sees need for it. If he is driving a car at speed of 54 km/h and the brakes cause deacc. of 6 m/s^2,find the distance travelled by car after he sees need to put brakes. (ANSWER GIVEN: 22 m) JUSTIFY your answer

Grade:11

2 Answers

Ramesh V
70 Points
12 years ago

The car is moving at speed of 54 km/hr i.e., 15 m/sec

that is initial speed after applying brakes is (u) = 15 m/sec

say  at point A,  he sees the need to put brakes

        at point B,  he puts brakes and then car starts deceleratin

         at point C,  car finally comes to rest

 

from A to B it moves with speed of 15m/sec in 0.2 sec

so distance covered is S1 = v.t = 15*0.2 = 3 mts

from B to C it moves with constant deceleration of -6 m/sec2 and initial speed of 15m/sec

Using relation V2 - u2 = 2*(-a)*S

S2 = V2/2*a = 15*15 /(2*6) =18.75 mts

the distance travelled by car after he sees need to put brakes is S1+S2

 = 21.75 mts  ( nearly 22 mts)

Kushagra Madhukar
askIITians Faculty 629 Points
one year ago
Dear student,
Please find the attached solution to your problem.
u = 54 km/hr = 15 m/s
v = 0 m/s
Distance covered before applying breaks, S1 = u.t = 15*0.2 = 3 mts
a =  – 6 m/s2
Using relation v2 - u2 = 2*(-a)*S
S2 = u2/2*a = 152 /(2*6) =18.75 mts
Hence distance covered, S = S1 + S2
                                        = 3 + 18.75
                                        = 21.75 mts (round off to 22 mts)
 
Hope it helps
Thanks and regards,
Kushagra

 

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