Question

A body starts with a velocity 4i-j+5km/s and moves with an acceleration of 7i+5j+6kms^-2. What is the velocity after 0.5s ?

Answer 1

Explanation:

v= u+at

u = 4i-j+5k

resultant of u = √{(4)²+(1)²+(5)²} =√(16+1+25= √42

a= 7i+5j+6k

resultant of a = √{(7)²+(5)²+(6)²= √(49+25+36)= √110

now

v= u+at

v=√42+√110 x 0.5

= √42 + 0.5√110

Answer 2

Answer:

The velocity of body after 0.5 s will be equal to $7.5\hat i-1.5\hat j+8\hat k \;\; Km/s$.

Explanation:

Given, the initial velocity of the body, $\vec u = (4\hat i-\hat j+5\hat k)Km/s$

The acceleration of that body is given as:

$\vec a = (7\hat i+5\hat j+6\hat k )Kms^{-2}$

Given,  the time, t = 0.5sec for which we have to calculate final velocity of the body.

We can use the first equation of motion to calculate velocity of body at time 't':

$\vec v =\vec u + \vec at$                                                                 ....................(1)

where u is initial velocity and a is acceleration and v is the final velocity at time 't'.

Substitute the values of u, a and t in equation (1) to calculate the velocity at time t.

$\vec v= (4\hat i-\hat j+5\hat k) + (7\hat i+5\hat j+6\hat k) \times (0.5)$

$\vec v= 4\hat i-\hat j+5\hat k + 3.5\hat i+2.5\hat j+3\hat k$

$\vec v= (7.5\hat i-1.5\hat j+8\hat k) Km/s$

Therefore, the velocity after 0.5 s will be equal to $7.5\hat i-1.5\hat j+8\hat k \;\; Km/s$.

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