Question

A body is moving with a velocity 4m/sec and acceleration 2m/sec find its velocity and distance travelled after 5 second.
{Note: how i solve this in exam}​

Answer 1

Answer:

Velocity = 14m/s

Distance = 45m

Explanation:

Initial Velocity(u) = 4m/s

Acceleration(a) = 2m/s²

Time(t) = 5seconds

Final Velocity(v) = ?

By 1st equation of motion

v = u + at

⟹ v = 4 + 2×5

⟹ v = 4 + 10

⟹ v = 14m/s

Therefore, its velocity is 14m/s.

Now, let's calculate distance

Initial Velocity(u) = 4m/s

Acceleration(a) = 2m/s²

Final Velocity(v) = 14m/s

Distance(s) = ?

By 3rd equation of motion

v² = u² + 2as

⟹ 14² = 4² + 2×2s

⟹ 14² = 4² + 4s

⟹ 196 = 16 + 4s

⟹ 4s + 16 = 196

⟹ 4s = 196 - 16

⟹ 4s = 180

⟹ s = 180/4

⟹ s = 45m

Therefore, its distance is 45m.

[In exam you can solve in this way if you want]