A) 4 m
B) 6 m
C) 6.8 m
D) 8 m
Correct Answer: A
Solution :
The position vector \[{{r}_{1}}\]of the particle at the instant t is given by \[{{r}_{1}}={{v}_{0}}t+1/2a{{t}^{2}}\] where \[{{v}_{0}}\]is the initial velocity and a is the constant acceleration We have \[{{v}_{0}}=2\hat{j}\] and \[a=2\hat{i}+4\hat{j}\]. Therefore, \[{{r}_{1}}=2\hat{j}\,t+(1/2)(2\hat{i}+4\hat{j}){{t}^{2}}={{t}^{2}}\hat{i}+(2t+2{{t}^{2}})\hat{j}\]. The above equation shows that the x-coordinate of the particle at time t is \[{{t}^{2}}\]and the y-coordinate is \[(2t+2{{t}^{2}})\] The time t at which the y-coordinate becomes 12 metres is given by \[2t+2{{t}^{2}}=12\]Or, \[2{{t}^{2}}+2t-12=0\] This gives \[t=[-\,2\pm \surd (4+96)]/4=2\] seconds, ignoring the negative time. Since the x-coordinate of the particle is \[{{t}^{2}},\]its value when the y-coordinate becomes 12 m (at time 2 seconds) is 4 m.
You need to login to perform this action.
You will be redirected in
3 sec
