Answer to Question #90430 in Mechanics | Relativity for AWESOME

Question

A particle moves with a uniform acceleration of 5m/s^2 in 10 seconds. what is the average distance covered in the last one second

Answer 1

Ehe equatins of the movement with unifom acceleration:

$$v=v_0+a*t$$

$$S=v_0*t+a*t^2/2$$

On the start of a movement:

$v_0=0, t=0$

Speed ($v_9$ ) and distance ($S_9$) after 9 seconds of movement:

$$v_9=a*t_9; S_9=a*t_9^2/2$$

Where $t_9=9s$ - time from movement start during 9 seconds

Distance ($S_{10}$​) after 10 seconds of movement:

$$S_{10}=a*t_{10}^2/2$$

Where $t_{10}=10s$ - time from bedinning during 10 seconds

The distance covered durind last decond:

$$S_x=S_{10}-S_9=a*t_{10}^2/2-a*t_9^2/2$$

Usind the numbers in SI:

$$S_x=47.5m$$

Answer: $S_x=47.5m$