The position vector of a particle is determined by the expression r -3 t 2-4 t 2-7 k- The distance traversed in first 10 sec isA- 500 mB- 300 mC- 100 mD- 150 m

The correct option is A 500 m r⃗=3t^2 î+4t^2 ĵ+7k̂ at t=0, r⃗_⃗1⃗=7k̂ at t=10 sec, r⃗_⃗2⃗=300 î+400ĵ+7k̂, Δ r=r⃗_2 r⃗_1=300 î+400ĵ |Δ r|=|r⃗_2 r⃗_1|=√((300)^ ...

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Standard XII

Physics

Addition and Subtraction in Unit Vector Notation

The position ...

Question

The position vector of a particle is determined by the expression
$\to r = 3 t^{2}^i + 4 t^{2}^j + 7^k$ . The distance traversed in first 10 sec is

500 m

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300 m

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150 m

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100 m

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Solution

The correct option is A
500 m

$\to r = 3 t^{2}^i + 4 t^{2}^j + 7^k$
at $t = 0,_{1} = 7^k$
at $t = 10 s e c,_{2} = 300^i + 400^j + 7^k,$
$- \to Δ r = {\to r}_{2} - {\to r}_{1} = 300^i + 400^j$
$| - \to Δ r | = | {\to r}_{2} - {\to r}_{1} | = \sqrt{(300)^{2} + (400)^{2}} = 500 m$ .

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The position vector of a particle is determined by the expression →r=3t2^i+4t2^j+7^k. The distance traversed in first 10 sec is

The correct option is A 500 m →r=3t2^i+4t2^j+7^k at t=0,→r1=7^k at t=10 sec,→r2=300^i+400^j+7^k, −→Δr=→r2−→r1=300^i+400^j |−→Δr|=|→r2−→r1|=√(300)2+(400)2=500 m.

The position vector of a particle is determined by the expression
$\to r = 3 t^{2}^i + 4 t^{2}^j + 7^k$ . The distance traversed in first 10 sec is

The correct option is A
500 m

$\to r = 3 t^{2}^i + 4 t^{2}^j + 7^k$
at $t = 0,_{1} = 7^k$
at $t = 10 s e c,_{2} = 300^i + 400^j + 7^k,$
$- \to Δ r = {\to r}_{2} - {\to r}_{1} = 300^i + 400^j$
$| - \to Δ r | = | {\to r}_{2} - {\to r}_{1} | = \sqrt{(300)^{2} + (400)^{2}} = 500 m$ .