Question

Derive Position-Time relation graphically.​

Answer 1

Answer:

s=area of rectangle OACD+area of triangle ABC

s=area of rectangle OACD+area of triangle ABCs=OA×OD+1/2×AC×BC

s=area of rectangle OACD+area of triangle ABCs=OA×OD+1/2×AC×BCs=u×t+1/2×t×(v-u)

s=area of rectangle OACD+area of triangle ABCs=OA×OD+1/2×AC×BCs=u×t+1/2×t×(v-u)s=ut +1/2×t×(v-u). --------eq 1

s=area of rectangle OACD+area of triangle ABCs=OA×OD+1/2×AC×BCs=u×t+1/2×t×(v-u)s=ut +1/2×t×(v-u). --------eq 1a=v-u/t

s=area of rectangle OACD+area of triangle ABCs=OA×OD+1/2×AC×BCs=u×t+1/2×t×(v-u)s=ut +1/2×t×(v-u). --------eq 1a=v-u/tat=v-u

s=area of rectangle OACD+area of triangle ABCs=OA×OD+1/2×AC×BCs=u×t+1/2×t×(v-u)s=ut +1/2×t×(v-u). --------eq 1a=v-u/tat=v-uput value of v-u in equation 1

s=area of rectangle OACD+area of triangle ABCs=OA×OD+1/2×AC×BCs=u×t+1/2×t×(v-u)s=ut +1/2×t×(v-u). --------eq 1a=v-u/tat=v-uput value of v-u in equation 1s=ut +1/2×t×at

s=area of rectangle OACD+area of triangle ABCs=OA×OD+1/2×AC×BCs=u×t+1/2×t×(v-u)s=ut +1/2×t×(v-u). --------eq 1a=v-u/tat=v-uput value of v-u in equation 1s=ut +1/2×t×ats=ut +1/2 at^2

s=area of rectangle OACD+area of triangle ABCs=OA×OD+1/2×AC×BCs=u×t+1/2×t×(v-u)s=ut +1/2×t×(v-u). --------eq 1a=v-u/tat=v-uput value of v-u in equation 1s=ut +1/2×t×ats=ut +1/2 at^2HENCE PROVE