A bird flies from Manhattan Beach to Brooklyn College along the Ocean Avenue. It

Question

A bird flies from Manhattan Beach to Brooklyn College along the Ocean Avenue. It flies with a constant speed for 4 km for 10 minutes arriving at Avenue Z. It then accelerates its speed from Avenue Z to Avenue M at which the birds traveling speed is 10m/s. From Avenue M to Brooklyn College, the bird's speed decreases from 10 m/s to zero when it arrives at Brooklyn College, traveling another 5km. What's the birds speed in between Manhattan Beach and Avenue Z? What's the value of acceleration from Avenue Z to M if the distance between the two avenues is 6km? A bird flies from Manhattan Beach to Brooklyn College along the Ocean Avenue. It flies with a constant speed for 4 km for 10 minutes arriving at Avenue Z. It then accelerates its speed from Avenue Z to Avenue M at which the birds traveling speed is 10m/s. From Avenue M to Brooklyn College, the bird's speed decreases from 10 m/s to zero when it arrives at Brooklyn College, traveling another 5km. What's the birds speed in between Manhattan Beach and Avenue Z? What's the value of acceleration from Avenue Z to M if the distance between the two avenues is 6km?

Explanation / Answer

=speed between Manhattan Beach to Avenue Z=initial speed=4000/600=40/6=6.67 m/s

at Z ,speed=10 m/s

let the accleration is a

distance=6000 m

using

v^2=u^2+2aS

a=v^2-u^2/2S=4.625*10^(-3) m/s^2

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