Question

A body rises vertically up to a height of 125m in 5s and then comes back at point of projection. Find: distance , displacement , average speed , average velocity of body

Answer 1

Given Distance S = 125 m

Time t = 5s

i) Distance travelled

Distance travelled = S + S = 2S

= 2 × 125 = 250m

$\fbox{ \rm{ \therefore \: Distance \: travelled = 250m}}$

ii) Displacement

Displacement = 0 (since final position is same as initial position)

iii) Average speed

$\rm Average \: speed = \frac{total \: distance \: travelled}{total \: time \: of \: journey}$

$\rm \implies\frac{2S}{2t} = \frac{2 \times 125m}{2 \times 5s}$

$\rm \implies \frac{250m}{10s} = 25 m{s}^{ - 1}$

$\rm \fbox{ \therefore average \: speed =25m {s}^{ - 1} }$

iv) Average Velocity

Average velocity = 0 (since displacement is zero)

Answer 2

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