Question

question number 2

Q2. A particle travels 10 m in first 5 seconds and 10 m in next 3 seconds. Assuming constant acceleration what is the distance travelled in next 2 seconds ?

(1) 8.3 m (2) 9.3 m (3) 10. 3 m (4) None of these

Answer 1

Dear Student ,
Here in this case let a be the acceleration and u be the initial velocity .

$$\begin{gathered} s=ut+\frac{1}{2}a{t}^2 \\ \Rightarrow 10=5u+\frac{25}{2}a.....\left(1\right) \\ Again,s=ut+\frac{1}{2}a{t}^2 \\ \Rightarrow 10=3u+\frac{9}{2}a......\left(2\right) \\ Solvingthistwoequationsweget, \\ 10u+25a=6u+9a \\ \Rightarrow 4u=-16a \\ \Rightarrow u=-4a \\ So,20=10u+25a \\ \Rightarrow 20=-40a+25a=-15a \\ \Rightarrow a=-\frac{20}{15}=-\frac{4}{3} \\ andu=\frac{16}{3} \\ Thereforethedis\tan cetravelledbytheparticlein2secondsis, \\ S=\left(\frac{16}{3}\times 2\right)-\left(\frac{1}{2}\times \left(-\frac{4}{3}\right)\times 4\right)=\frac{32}{3}-\frac{8}{3}=\frac{24}{3}=8m \end{gathered}$$
Regards