Question

18. A particle has initial velocity (2+4j) m/s and
retardation (4i +8j) m/s2. The distance travelled
by particle from t = 0 to t = 1 s is
(1) 15 m
(2) T6 m
(3) V7 m
(4) data insufficient​

Answer 1

We have

$a_{x} = - 4 \\ a_{y} = - 7 \\ u_{x} = 2 \\ u_{y} = 4 \\ v_{x} = 2 - 4 \\ v_{y} = 4 - 8t$

According To Question

$Both \: v_{x} \: and \: v_{y} \: are \: 0 \: at \: t = \frac{1}{2} s$

After this particle returns to initial point

$S_{x} \: at \: t = \frac{1}{2} s \\ S_{x} = 2 \times \frac{1}{2} - \frac{1}{2} \times 4 \times \frac{1}{4} \\ S_{x} = \frac{ \cancel2}{ \cancel2} - \frac{1}{2} \times \frac{ \cancel4}{ \cancel4} \\ S_{x} = 1 - \frac{1}{2} \\ \boxed{S_{x} = \frac{1}{2} }$

$\bold{S_{y} \: at \: t = \frac{1}{2}s } \\ S_{y} = 4 \times \frac{1}{2} - \frac{1}{2} \times 8 \times \frac{1}{4} \\ S_{y} = \frac{ \cancel4}{ \cancel2} - \frac{ \cancel8}{ \cancel8} \\ S_{y} = 2 - 1 \\ \boxed{S_{y} = 1}$

Distance Travelled, D

$D = 2 \times \sqrt{ {S_{x}}^{2} + {S_{y}}^{2} } \\ D = 2 \sqrt{ \frac{1}{4} + 1 } \\ D = 2 \sqrt{ \frac{5}{4} } \\ D = \frac{ \cancel2}{ \cancel2} \sqrt{5} \\ D = \sqrt{5}$

Answer 2

Answer:

4. is the correct answer