Question

A body starts with a velocity (2i+3j+11k) m/s and moves with an acceleration (5i+5j-5k)m/s^2 what is its velocity after 0.2sec

Answer 1

Answer:

the velocity is 3i + 4j + 10k

Explanation:

using the first equation of motion

v = u + at

we are given u , a , t and to find v

v = 2i + 3j + 11k + 0.2( 5i + 5j - 5k )

v = 2i + 3j + 11k + i + j - k

v = 3i + 4j + 10k

Answer 2

Given :

Initial velocity (u) = 2i + 3j + 11k

Acceleration (a) = 5i + 5j - 5k

To Find :

Final velocity (v)

Solution :

We know, acceleration is the rate of change of velocity.

    Acceleration (a)     =  $\frac{v - u}{Time (T)}$

⇒   5i + 5j - 5k            =  $\frac{v - u}{0.2}$

⇒  0.2×(5i + 5j - 5k)    =  v - u

⇒  (i + j - k)                  =  v - (2i + 3j + 11k)

⇒      v                         =  (i + j - k) + (2i + 3j + 11k)

⇒      v                         =  3i + 4j + 10k

Therefore, the velocity after 0.2 sec is 3i + 4j + 10k