Question

A particle is projected upwards with velocity' of 20 ms^-1. 

Calculate its distance travelled and its displacement after 3.0 s. (g = 10ms^-2).

Answer 1

Given: u = -20 ms^-1

a = +g = +10 ms^-2

∴ Time taken upto highest point where velocity v = 0.

∴ Using equation v = u + at

0 = -20 - gt₁ = -20 - 10t₁

or 10t₁ = 20

∴ t₁ = \(\frac{20}{10}\) = 2 s

∴ t₂ = t - t₁ = 3 - 2 = 1 s

h₁ = ut₁ + \(\frac{1}{2}\) gt₁² = -20 x 2 + \(\frac{1}{2}\) x 10 x 4

= -40 + 20

= -20 m

Now again using S = ut² + \(\frac{1}{2}\) at₂²

h₂ = 0 + \(\frac{1}{2}\) x 10 x 1 = 5 m

∴ Total distance |h₁| + h₂ = 20 + 5 = 25 cm

Displacement |h₁| - h₂ = 20 - 5 = 15 cm

Based on average speed and average velocity