Q123 / 160 · TS EAMCET 2024 (10 May Shift 1) · Physics
A body moving with uniform acceleration, travels a distance of 25 m in the fourth second and 37 m in the sixth second. The distance covered by the body in the next two seconds is
- A63 m
- B84 m
- C49 m
- Correct answerD92 m
Solution
For uniformly accelerated motion,
$\begin{aligned}
& \mathrm{S}_{\mathrm{n}}=\mathrm{u}+(2 \mathrm{n}-1) \frac{\mathrm{a}}{2} \\
& \therefore \quad 25=\mathrm{u}+(2 \times 4-1) \frac{\mathrm{a}}{2}=\mathrm{u}+\frac{7}{2} \mathrm{a}
\end{aligned}$
Again, $37=u+(2 \times 6-1) \frac{a}{2}=u+\frac{11}{}$ From eq(i) and (ii), we get $\begin{aligned} & \mathrm{a}=6 \mathrm{~ms}^{-2}, \mathrm{u}=4 \mathrm{~km} \mathrm{~s}^{-1} \\ & \therefore \quad \text { At time, } \mathrm{t}=6 \mathrm{~s} \\ & \mathrm{v}=\mathrm{u}+\mathrm{at}=4+6 \times 6=40 \mathrm{~ms}^{-1} \\ & \therefore \quad \mathrm{~S}=\mathrm{vt}_1+\frac{1}{2} \mathrm{at}_1^2=40 \times 2+\frac{1}{2} \times 6 \times 2 \times 2 \\ & =92 \mathrm{~m} \end{aligned}$
Again, $37=u+(2 \times 6-1) \frac{a}{2}=u+\frac{11}{}$ From eq(i) and (ii), we get $\begin{aligned} & \mathrm{a}=6 \mathrm{~ms}^{-2}, \mathrm{u}=4 \mathrm{~km} \mathrm{~s}^{-1} \\ & \therefore \quad \text { At time, } \mathrm{t}=6 \mathrm{~s} \\ & \mathrm{v}=\mathrm{u}+\mathrm{at}=4+6 \times 6=40 \mathrm{~ms}^{-1} \\ & \therefore \quad \mathrm{~S}=\mathrm{vt}_1+\frac{1}{2} \mathrm{at}_1^2=40 \times 2+\frac{1}{2} \times 6 \times 2 \times 2 \\ & =92 \mathrm{~m} \end{aligned}$
Hence, the correct answer is option: 'D'.