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A horse canters away from its trainer in a straight line,moving 160m away in 17s. It then turns abruptly and gallops halfwayback in 6.8s. Calculate the (a) its average speed and (b) its average velocity for the entire trip, using "away from the trainer"as the positive direction.

Respuesta :

Answer:

a)10.08 m/s

b)3.36 m/s

Explanation:

Given that

distance s₁= 160 m

time t₁= 17 s

distance s₂= 80 m

time t₂ = 6.8 s

As we know that

Total distance = Total time x Average speed

s₁ + s₂ = (t₁ +  t₂ ) Vavg

[tex]V_{avg}=\dfrac{160+80}{17+6.8}\ m/s[/tex]

Vavg=10.08 m/s

Total displacement = Total time x Average velocity

s₁ - s₂ = (t₁ +  t₂ ) Vavg

[tex]V_{avg}=\dfrac{160-80}{17+6.8}\ m/s[/tex]

Vavg=3.36 m/s

a)10.08 m/s

b)3.36 m/s

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