An object is accelerating uniformly from 8.0 m/s to 16 m/s in 10 seconds. How far does the object travel in 10 seconds
acceleration, a = (16 - 8)/10 m/s2 = 0.8 m/s2.
Use v2 = u2 + 2as where v = final velocity, u = initial velocity, a = acceleration, s = distance.
Rearrange as:
s = (v2 - u2)/(2a)
so
s = (162 - 82)/(2*0.8) m
$${\mathtt{s}} = {\frac{\left({{\mathtt{16}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{{\mathtt{8}}}^{{\mathtt{2}}}\right)}{{\mathtt{1.6}}}} \Rightarrow {\mathtt{s}} = {\mathtt{120}}$$
s = 120m
Check:
You could also use s = ut + (1/2)at2 where t is time.
s = 8*10 + (1/2)*0.8*102m
$${\mathtt{s}} = {\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{10}}{\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}{\mathtt{0.8}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{2}}} \Rightarrow {\mathtt{s}} = {\mathtt{120}}$$
s = 120m
.
acceleration, a = (16 - 8)/10 m/s2 = 0.8 m/s2.
Use v2 = u2 + 2as where v = final velocity, u = initial velocity, a = acceleration, s = distance.
Rearrange as:
s = (v2 - u2)/(2a)
so
s = (162 - 82)/(2*0.8) m
$${\mathtt{s}} = {\frac{\left({{\mathtt{16}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{{\mathtt{8}}}^{{\mathtt{2}}}\right)}{{\mathtt{1.6}}}} \Rightarrow {\mathtt{s}} = {\mathtt{120}}$$
s = 120m
Check:
You could also use s = ut + (1/2)at2 where t is time.
s = 8*10 + (1/2)*0.8*102m
$${\mathtt{s}} = {\mathtt{8}}{\mathtt{\,\times\,}}{\mathtt{10}}{\mathtt{\,\small\textbf+\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}{\mathtt{0.8}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{{\mathtt{2}}} \Rightarrow {\mathtt{s}} = {\mathtt{120}}$$
s = 120m
.