Question

A driver in a bus moving with 40 m/s sees a girl at 25m distance. If he retards bus at 5 m/s2 , can he save the girl or not? solve using equation of motion formula plzzzzz with whole process those who spam will be blocked forever so plz wwrite only if you know answers like i dont know,question not clear will not be accepted and taken strictly into action

Answer 1

Acceleration, a=

Time

v

final

−v

initial

Therefore, −2m/s

2

=

5s

v

final

−20m/s

v

final

=10m/s

Answer 2

A car is moving with speed 20m/s. Suddenly the driver sees the sign of danger at a distance of 50m; after a certain reaction time t

he applies brakes to cause deceleration 5m/s2

. What is the maximum allowable reaction time t

to avoid accident and distance travelled by the during reaction time?

ANSWER

Given,

Speed=20m/s,d=50m,retardation=5m/s2

Let the ppossible reaction time be t sec

So the car moves 20tm And applied breaks with retardation 5m/s2

So, the total distance under retardation is ≤50m

Thus the distance under the application of retardation =

2×520 2−02

=s

′

⇒s =40m

Therefore, 20t +40≤=50⇒20t s≤50−40⇒20t s=10⇒ts ≤ 21≤0.5s

HOPE IT HELPS UH DEAR