Question

PLEAS HELP ME IN THE BELLOW ATTCHMENT

f61dda5abc64fd9bda15aea38decfbb8.pdf

Answer 1

Answer: (a) i.AB = acceleration

ii.BD = constant/ steady speed

iii.DE= deceleration

(b) 20$km/h^{2}$  , 0 $km/h^{2}$, -20 $km/h^{2}$

(c) 160 $km^{2}$

(d) 20 km/h

Explanation:

(a)On speed-or-velocity/time graph :

• AB positive slope represents increasing speed (acceleration)

• a line like BD which is just a straight line represents constant/steady speed

• DE is the opposite of AB -it goes down- it's a negative slope so it's decreasing speed (deceleration)

(b) to calculate acceleration you can use the following formula

$\frac{final velocity - initial velocity}{time}$

here is all the working for answers in (b)

$\frac{40 - 0}{2}$ = 20$km/h^{2}$ ,  $\frac{40-40}{4}$ = 0$km/h^{2}$ , $\frac{0-40}{2}$ = -20$km/h^{2}$$\frac{160}{8} = 20$

(c) to calculate the distance in a velocity/time graph you have to calculate the area under the graph

so: $\frac{1}{2} *2 *40 + 2*40 +\frac{1}{2} *2 *40$    (please note that the '*' is meant to be the multiply sign)

(d) to calculate the average speed you have to divide the total distance by the total time taken. So in this case:

(we will use the distance we determined in (c)) and divide by 8 the 'hours'

$\frac{160}{8} = 20$