Question

A body covers a distance of 4 m in 3rd second and 12 m in 5th second. If the motion is uniformly accelerated, how far will it travel in the next 3 seconds? 

1) 60 m 

2) 70 m 

3) 80 m 

4) 100 m

Answer 1

Correct option is 1) 60 m

\( S_3=u+\frac{a}{2}(2 \times 3-1)=4 \text { or } u+\frac{5}{2} a=4 \)

\(S_5=u+\frac{a}{2}(2 \times 5-1)=12 \text { or } u+\frac{9}{2} a=1\)

\(\text { On solving, } \mathrm{u}=-6 \mathrm{~ms}^{-1}, \mathrm{a}=4 \mathrm{~ms}^{-2} \)

\(\text { Distance travelled in next } 3 \text { seconds }=S_8-S_5\)

\(\left[-6 \times 8 + \frac {1}{2} \times 4 \times (8)^2 \right] - \left[-6 \times 5 + \frac {1}{2} \times 4 \times (5)^2 \right]\)

80 - 20 = 60 cm