Question

please guys answer me fast and correctly
I'll mark him or her as the brainiest
and don't answer if you don't know​

Answer 1

Number of motion shown by AB, BC, CD

• AB showing uniformly accelerated motion
• BC showing uniformly constant motion
• CD showing uniformly deaccelerated motion

Value of acceleration=100/49ms^-2

Value of acceleration=100/49ms^-2Value of distance= 200M

Explanation:

To calculate distance from V-T graph simply find area of graph.

$2 \times 4 \times 20 \times \frac{1}{2} = 80m$

$6 \times 20 = 120m$

$\large \: Total \: distance=200m$

To calculate acceleration put the values in equation 2 of motion

$s = ut + \frac{1}{2} \times a \times {t}^{2}$

$200 = 0 \times 14+ \frac{1}{2} \times a \times {14}^{2}$

$\huge \: a = \frac{200}{98} = \frac{100}{49}$

Formula used

• $s = ut + \frac{1}{2} \times a \times {t}^{2}$
• $area \: of \: triangle \: = \frac{1}{2} \times b \times h$
• $area \: of \: rectangle= l \times b$