Question

When a constant braking force is applied to a vehicle moving at speed v, the distance d moved by the vehicle in coming to rest is given by the expression d = kv^2 where k is a constant. When d is measured in metres and v is measured in metres per second, the constant has a value of k1. What is the value of the constant when the distance is measured in metres, and the speed is measured in kilometres per hour?
Update: MCQ options A 0.0772 k B 0.278 k C 3.60 k D 13.0 k
PLEASE HELP ASAP I HAVE A HUGE PHYSICS MOCK EXAM TOMMORROW!!

Answer 1

d=kv^2
d in meters v in km/h
so ,
v^2 unit =km^2/h^2
=(1000 m)^2/(3600sec)^2
=(1/3.6)^2 m^2/sec^2
=(1/12.96) m^2/sec^2
d m=kv^20.0772 m^2 /sec^2===========(1)

question given
d m=k1v^2 m^2/sec^2=========(2)
compared both of this ,
k1=k (0.0772)
k=1/(0.772) k1
=12.96 k1
=13 k1
hence option (D) is correct answer.

Answer 2

Answer:

The answer is A!!

Explanation:

d = kv2

When speed v is in ‘m/s’ and distance d is in ‘m’, the value of k is ‘k1’.

k1 = d / v2

We want to know the value of k when v is in ‘km/h’ and d in ‘m’.

Converting m/s to km/h

1 m/s means that in 1s - - -> 1m

1 h = 3600 s

1s - - -> 1m

1h - - > 3600 m = 3.6 km

Thus, 1 m/.s is equal to 3.6 km/h. Let the new value of speed = v2.

v2 = 3.6 v

New value of k = d / v22 = d / (3.6v)2 = (1 / 3.62) (d / v2)

But d / v2 = k1  

New value of k = (1 / 3.62) k1 = 0.0772 k1