Answer to Question #148242 in Classical Mechanics for David Okechukwu

Question

A lion starts 26m away from a clueless peter and charges towards him at a constant velocity of 50km/h,it takes peter 1 second to react to the lion,turn around and begin running to his vehicle at a velocity of +5m/s. Peter's Bentley is parked 6m away from him on the same axis as the lion's charge.
If peter escapes,how far is he from the lion..it if peter is caught,how far behind is the lion

Answer 1

Answer

Changing speed in m/sec

50km/h=13.88m/sec

Distance travelled in 1sec=13.88m

So time taken by Peter t₁=6/5=1.2sec

Time taken by lion t₂=$\frac{12.1+6}{13.9}=1.30sec$

So t₁<t₂ so peter will escape.

Now distance travelled by lion in 1.2sec

=13.9$\times1.2$ =16.7m

So Peter behind the lion

=18.1-1.67=1.4m