Question

the position vector of a particle is determined by the expression r= 3t^{2} i+4t^{2}+7k.the distance traversed in first 10 sec is ???
a)500m b)300m c)150m d)100m...
please explain your answer with clear solution

Answer 1

maybe U asked Displacement. Calculating distance will be cumbersome as it is some complex parabola

r (i)= Initial position vector
r (f) = Final position vector

Since r is the finction of time.

r(i)=r(0)=7 k^   
r(f)=r(10)=3*10² i^ + 4*(10)² j^ +7 k^

Now ,
Displacement vector= r(f)-r(i)
D= |300 i^ + 400j^ +(7-7)k^ |
 
$\sqrt{ 300^{2} + 400^{2} }$

=500 

(a) is correct !!

Answer 2

r (i)= Initial position vector

r (f) = Final position vector

Since r is the finction of time.

r(i)=r(0)=7 k^   

r(f)=r(10)=3*10² i^ + 4*(10)² j^ +7 k^

Now ,

Displacement vector= r(f)-r(i)

D= |300 i^ + 400j^ +(7-7)k^ |

 

=500 

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