Question

A body is moving with uniform acceleration and describes 75 m in the 6th second and 115 m in the 11th second. Show that its moves 155 m during the 16th second.

Answer 1

Step 1: Given data

S_6^th = 75 meters (m)

S_11^th = 7 meters (m)

Let the Initial velocity be u and uniform acceleration be a.

Step 2: Forming an equation for u

$$\begin{gathered} {\mathrm{S}}_{6^{\mathrm{th}}}={\mathrm{S}}_6-{\mathrm{S}}_5=75\mathrm{m} \\ \left({\mathrm{ut}}_6+\frac{1}{2}\mathrm{a}{{\mathrm{t}}_6}^2\right)-\left({\mathrm{ut}}_5+\frac{1}{2}\mathrm{a}{{\mathrm{t}}_5}^2\right)=75\mathrm{m} \\ \left(6\mathrm{u}+\frac{36\mathrm{a}}{2}\right)-\left(5\mathrm{u}+\frac{25\mathrm{a}}{2}\right)=75\mathrm{m} \\ \mathrm{u}+\frac{(36\mathrm{a}-25\mathrm{a})}{2}=75\mathrm{m} \\ \mathrm{u}=75-\frac{11\mathrm{a}}{2}----\left(1\right) \end{gathered}$$

Step 3: Finding the value of u and a

$$\begin{gathered} {\mathrm{S}}_{{11}^{\mathrm{th}}}=115\mathrm{m} \\ {\mathrm{S}}_{11}-{\mathrm{S}}_{10}=115\mathrm{m} \\ \left({\mathrm{ut}}_{11}+\frac{1}{2}\mathrm{a}{{\mathrm{t}}_{11}}^2\right)-\left({\mathrm{ut}}_{10}+\frac{1}{2}\mathrm{a}{{\mathrm{t}}_{10}}^2\right)=115\mathrm{m} \\ \left(11\mathrm{u}+\frac{121\mathrm{a}}{2}\right)-\left(10\mathrm{u}+\frac{100\mathrm{a}}{2}\right)=115\mathrm{m} \\ \mathrm{u}+\frac{121\mathrm{a}-100\mathrm{a}}{2}=115\mathrm{m} \\ \mathrm{from}\left(1\right) \\ 75-\frac{11\mathrm{a}}{2}+\frac{21\mathrm{a}}{2}=115\mathrm{m} \\ \frac{21\mathrm{a}-11\mathrm{a}}{2}=115-75 \\ 5\mathrm{a}=40 \\ \mathrm{a}=8\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Putting}\mathrm{value}\mathrm{of}\mathrm{a}\mathrm{in}\mathrm{equation}\left(1\right); \\ \mathrm{u}=75-\frac{11\left(8\right)}{2} \\ \mathrm{u}=75-44=31\mathrm{m}/\mathrm{s} \end{gathered}$$

Step 4: Finding the value of distance at the 16^th second:

$$\begin{gathered} {\mathrm{S}}_{{16}^{\mathrm{th}}}={\mathrm{S}}_{16}-{\mathrm{S}}_{15} \\ =\left({\mathrm{ut}}_{16}+\frac{1}{2}\mathrm{a}{{\mathrm{t}}_{16}}^2\right)-\left({\mathrm{ut}}_{15}+\frac{1}{2}\mathrm{a}{{\mathrm{t}}_{15}}^2\right) \\ \mathrm{putting}\mathrm{u}=31\mathrm{m}/\mathrm{s}\mathrm{and}\mathrm{a}=8\mathrm{m}/{\mathrm{s}}^2 \\ =\left[31\left(16\right)+\frac{1}{2}\left(8\right){\left(16\right)}^2\right]-\left[31\left(15\right)+\frac{1}{2}\left(8\right){\left(15\right)}^2\right] \\ =\left[496+4\left(256\right)\right]-\left[465+4\left(225\right)\right] \\ =496+1024-465-900 \\ =155\mathrm{metres}\left(\mathrm{m}\right) \end{gathered}$$

Hence, we proved that the body moves 155 m during the 16^th second.