An object is moving with a velocity 4m/s.If it is subjected to an acceleration of 2m/s^2?

Question

calculate its velocity and distance travelled after 5 second.

(m/s->meter per second).

Answer 1

Final Velocity: Initial Velocity + (acceleration*time)

Thus, 4 + (2*5)= 14 m/s.

Distance: Average Velocity * Time

Average Velocity=Final Velocity+Initial Velocity/2

Thus, A.V= 18/2=9 m/s.

Further, Distance= 9*5= 45 m.

Answer 2

Initial velocity u = 4 m/s

Acceleration a = 2 m/s^2

Time t = 5 s

Final velocity v = ?

Distance travelled s = ?

v = u + at = 4 + 2*5 = 4 + 10 = 14 m/s

s = ut + 1/2 at^2 = 4*5 + 1/2*2*5^2 = 20 + 25 = 45 m

Ans: velocity = 14 m/s

Distance = 45 m

Answer 3

what the h**l are you talking about? school hasn't started yet

Answer 4

initial velocity, u = 4 m/s

time taken, t = 5 s

acceleration, a = 2m/s^2

final velocity, v = u + at

v = 4 + (2) (5) = 14 m/s (ans)

Using s = ut + 0.5 at^2

s = 4 X 5 + 0.5 X 2 X 5^2

s = 45 m (ans)