Please give solution..
Dear student
You can refer to the above solution for similarquery.
Let policeman catch the driver after time t, distance travelled by both of them is same which is 1.5 km = 1500 m.
Let v be the speed of the speeder.
v = distance/ time
v = 1500/t
Initial speed of the policeman is 0.
final velocity after 12 s is 150 km/h = 125/3 m/s
v = u + at
125/3 = 0+ 12a
a = 125/36 m/s2
distance travelled in 12 s is:
s = ut + 1/2at2
s = 0 + (1/2)×(125/36)×12×12
s = 250 m
Distance travelled in time t = distance travelled in 12 s + distance travelled in time t - 12 - 2 seconds moving with speed 125/3 m/s
because 2 second is the respond time of the policeman.
1500 = 250 + (125/3)(t -12 - 2)
t = 44 s
Now velocity of the speeder
v = 1500/44
v = 34.09 m/s
v = 122.7 km/h
REGARDS
You can refer to the above solution for similarquery.
Let policeman catch the driver after time t, distance travelled by both of them is same which is 1.5 km = 1500 m.
Let v be the speed of the speeder.
v = distance/ time
v = 1500/t
Initial speed of the policeman is 0.
final velocity after 12 s is 150 km/h = 125/3 m/s
v = u + at
125/3 = 0+ 12a
a = 125/36 m/s2
distance travelled in 12 s is:
s = ut + 1/2at2
s = 0 + (1/2)×(125/36)×12×12
s = 250 m
Distance travelled in time t = distance travelled in 12 s + distance travelled in time t - 12 - 2 seconds moving with speed 125/3 m/s
because 2 second is the respond time of the policeman.
1500 = 250 + (125/3)(t -12 - 2)
t = 44 s
Now velocity of the speeder
v = 1500/44
v = 34.09 m/s
v = 122.7 km/h
REGARDS