Question

19. The velocity of an object at t = 0 is V. =-4; ms. It moves in a plane with constant
acceleration = (31+8) m/s2. What is its velocity after 1 s?​

Answer 1

Answer:

Explanation:

v = 3i + 4j m/s

|v| = 5 m/s

◆ Explaination-

# Given-

Initial velocity u = -4j m/s

Time interval t = 1 s

Acceleration a = 3i + 8j m/s^2

# Solution-

Using Newton's 1st kinematics eqn,

v = u + at

v = -4j + (3i+8j)1

v = 3i + 4j m/s

Magnitude of velocity is -

|v| = √(3^2 + 4^2)

|v| = √(9 + 16)

|v| = √25

|v| = 5 m/s

Therefore, velocity is 3i + 4j m/s with magnitude 5 m/s.

Hope this helps....

Answer 2

Answer:

KE=1/2mv²

mass=0.3kg

velocity,=5

KE=1/2×0.3×5×5=3.75J