question_answer

A particle of mass 10 kg starts from point A, with an initial velocity of 3m/s towards negative x-axis, it has been acted by a force of 10 N towards positive x-axis. Find the distance travelled by particle in 4 s.

A)  4 m                            

B)  0.5 m

C)  4.5m                          

D)  5m

Correct Answer: D

Solution :

This is a question based on the concept that when acceleration and initial velocity are in opposite direction then particle comes to rest momentarily at some instant and after that direction of velocity becomes same as that of acceleration. Up to this instant distance and displacement are equal. \[a=\frac{F}{m}\] \[=\frac{10}{10}=1m/{{s}^{2}}\] \[\leftarrow =3m/s,t=0\] \[\xleftarrow[{}]{{}}{{s}_{1}}\xrightarrow[{}]{{}}\] Let velocity becomes zero after t second, then applying v = u + at \[\Rightarrow \]\[0=3-1\times t\]\[\Rightarrow \]\[t=3s\] Distance travelled in 3 s = Displacement in 3 s \[\Rightarrow \]\[{{s}_{1}}=ut+\frac{1}{2}a{{t}^{2}}=3\times 3-\frac{1}{2}\times 1\times {{3}^{2}}\]\[=\frac{9}{2}m=4.5m\] [The above equation is a vector equation and gives us dish placement cement and not distance] D is placement of particle in 4 s \[s=ut+\frac{1}{2}a{{t}^{2}}\] \[=3\times 4-\frac{1}{2}\times 1\times {{4}^{2}}\] \[=12-8=4m\]From the diagram, distance travelled in 4 s\[=AB+BC=4.5+0.5=5m\]