A police jeep is chasing a culprit going on a motorbike. The motorbike crosses a turning at a speed of 72 km/h. the jeep follows it at a speed of 90 km/h, crossing the turning ten seconds later than the bike. Assuming that they travel at constant speeds, how far from the turning will the jeep catch up with the bike?

Deepak Patra
9 years ago
Sol. VP = 90 km/h = 25 m/s. VC = 72 km/h = 20 m/s. In 10 sec culprit reaches at point B from A. Distance converted by culprit S = vt = 20 × 10 = 200 m. At time t = 10 sec the police jeep is 200 m behind the culprit Time = s/v = 200 / 5 = 40 s. (Relative velocity is considered). In 40 s the police jeep will move from A to a distance S, where S = vt = 25 × 40 = 1000 m = 1.0 km away. ∴ The jeep will catch up with the bike, 1 km far from the turning.
Pratham
13 Points
4 years ago
Culprits speed 20 m/s , police's jeep speed 25 m/s
Now dist b/w them would be equal to velocity of the culprits bike * time he took to drive the bike =V*T=20*10=200
Now let dist b/w the culprit and the final mmeting position be x so time taken by culprit = time taken by police to arrive at final position =t1=t2= 200+x/25=x/20= x= 800m total dist 800+ initial gap 200.= 1000m =1 k m plz appreciate it
3 years ago
Dear student,

VP = 90 km/h = 25 m/s.
VC = 72 km/h = 20 m/s.
In 10 sec culprit reaches at point B from A.
Distance converted by culprit(B → A), S = vt = 20 × 10 = 200 m.
At time t = 10 sec the police jeep is 200 m behind the culprit
Relative speed of jeep w.r.t culprit = 25 – 20 = 5 m/s
Time in which the gap is covered by the police = s/v = 200 / 5 = 40 s.
In 40 s the police jeep will move from A to a distance S, where
S = vt = 25 × 40 = 1000 m = 1.0 km away.
∴ The jeep will catch up with the bike 1 km far from the turning.

Thanks and regards,
Kushagra