Question

An object uniformly accelerates from rest and reaches a velocity of 122.0 km/h north in 10.5 s. what was the average velocity of the object?

Answer 1

An object uniformly accelerates from rest and reaches a velocity of 122.0 km/h north in 10.5 s. what was the average velocity of the object?

The formula for average velocity:

$$v _ {a} = \frac {S _ {t o t a l}}{t _ {t o t a l}}$$

Total distance:

$$S _ {t o t a l} = \frac {a t _ {t o t a l} ^ {2}}{2}$$

By definition, acceleration is (initial velocity = 0):

$$a = \frac {v _ {f i n a l}}{t _ {t o t a l}}$$

Thus:

$$S _ {t o t a l} = \frac {\frac {v _ {f i n a l}}{t _ {t o t a l}} t _ {t o t a l} ^ {2}}{2} = \frac {t _ {t o t a l} v _ {f i n a l}}{2}$$

And finally:

$$v _ {a} = \frac {t _ {t o t a l} v _ {f i n a l}}{2 t _ {t o t a l}} = \frac {v _ {f i n a l}}{2}$$ $$v _ {a} = \frac {1 2 2 k m / h}{2} = 6 1 k m / h$$

Answer: $v_{a} = 61km / h$