Question

20 POINTS

Answer 1

Answer:

c

Explanation:

Answer 2

Answer:

10m

Explanation:

let's take the acceleration as a constant throughout the complete motion...

therefore first let's find the acceleration

$s = ut + \frac{1}{2} a {t}^{2} \\ 2 = \frac{1}{2} a {1}^{2} \\ a = 4m {s}^{ - 2}$

then we have to find v1

apply V = u + at

v = 4×1

= 4ms^-1

lets find the distance travel by the object DURING THE TIME INTERVAL 1-2

$s = ut + \frac{1}{2} a {t}^{2} \\ s = 4 \times 1 + \frac{1}{2} \times 4 \times {1}^{2} \\ s= 6m$

then let's find the V2

${v}^{2} = {u}^{2} + 2as \\ {v}^{2} = {4 }^{2} + 2 \times 4 \times 6 \\ {v}^{2} = 64 \\ v = \sqrt{64} = 8m {s}^{ - 1}$

then let's find the distance travel by the object during time interval 2s to 3s

$s = ut + \frac{1}{2} a {t}^{2} \\ s = 8 \times 1 + \frac{1}{2} \times 4 \times {1}^{2} \\ s = 10m$