A) 24 m
B) 40 m
C) 56 m
D) 16 m
Correct Answer: C
Solution :
When particle comes to rest, \[\text{V=0=}\frac{\text{dx}}{\text{dt}}\text{=}\frac{\text{d}}{\text{dt}}\left( 40+12\text{t}-{{\text{t}}^{3}} \right)\] \[\Rightarrow 12-3{{\text{t}}^{\text{2}}}=0\] \[\Rightarrow {{\text{t}}^{\text{2}}}=\frac{12}{3}=4\text{ }\therefore \text{t=2 sec}\] Therefore distance travelled by particle before coming to rest, \[x=40+12t-{{t}^{3}}=40+12\times 2-{{\left( 2 \right)}^{3}}=56\text{m}\]
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