Question

Practice Problems

Answer 1

(1) The distance traveled by Thomas is 190 m

(2) The displacement of Thomas is 100 m forward

(3) The velocity of Thomas is 0.056 m/s

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(1) Distance is the sum of the total path covered.

first distance to cafe = 100 m

second distance to library = 45 m

third distance back to the cafe = 45 m

The total distance = 100 + 45 + 45 = 190 m

The distance traveled by Thomas is 190 m

(2) Displacement is the change in the position of an object;

Initial displacement forward = (100 + 45) m

Final displacement backward = 45 m

The total displacement = 145 m - 45 m = 100 m forward

The displacement of Thomas is 100 m forward

(3) The velocity of Thomas is calculated as the change in displacement per change in time.

the change in displacement = 100 m

the time of motion, t = 30 mins = 30 x 60 = 1800 s

$v = \frac{\Delta x}{\Delta t} = \frac{100 \ m}{1800 \ s} = 0.056 \ m/s$

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