Question

Question #a18d2

Answer 1

`"speed : "19.30 " "m/s`

`"displacement : "59" "meters`

Explanation:

`"the slope of graph is equal to acceleration"`

`tan alpha=3.00=(v_f-4.30)/5`

`3*5=v_f-4.30`

`15=v_f-4.30`

`v_f=15+4.30`

`v_f=19.30" "m/s`

`"area under graph is equal to displacement "`

`delta x=(4.30+19.30)/2*5`

`Delta x=(23.60)/2*5`

`Delta x=59" "meters`

`"or..."`

`v_f=v_i+a*t`

`v_f:"final speed"`

`a:"acceleration"`

`t:"time elapsed"`

`v_f=4.30+3*5`

`v_f=4.30+15`

`v_f=19.30" "m/s`

`Delta x=v_i*t+1/2*a*t^2`

`Delta x=4.30*5+1/2*3*5^2`

`Delta x=21.5+37.5`

`Delta x=59" "meters`