JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
A 4 kg mass moves under the influence of a force \(\vec{F}=\left(4 t^3 \hat{i}-3 t \hat{j}\right) N\) where \(t\) is the time in second. If mass starts from origin at \(t=0\), the velocity and position after \(t=2 s\) will be :
- A \(\vec{v}=3 \hat{i}+\frac{3}{2} ~\hat{j} \vec{r}=\frac{6}{5} \hat{i}+\hat{j}\)
- B \(\vec{v}=4 \hat{i}-\frac{3}{2} ~\hat{j} \vec{r}=\frac{8}{5} \hat{i}-\hat{j}\)
- C \(\vec{v}=4 \hat{i}+\frac{5}{2}~ \hat{j} \vec{r}=\frac{8}{5} \hat{i}+2 \hat{j}\)
- D \(\vec{v}=4 \hat{i}-\frac{3}{2} ~\hat{j} \vec{r}=\frac{6}{5} \hat{i}-\hat{j}\)
Answer & Solution
Correct Answer
(B) \(\vec{v}=4 \hat{i}-\frac{3}{2} ~\hat{j} \vec{r}=\frac{8}{5} \hat{i}-\hat{j}\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ F }=4 t^3 \hat{ i }-3 t \hat{ j }\) \(\overrightarrow{ a }=\frac{\overrightarrow{ F }}{ m }= t ^3 \hat{ i }-\frac{3}{4} \hat{ tj }\) \(a_x=t^3\) \(\frac{ dv _{ x }}{ dt }= t ^3\) \(\int_{ v _{ x }=0}^{ v _{ x _2}} d v_{ x }=\int_{ t =0}^{ t =2} t^3 d t\)…
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