Question

An object is moving with an initial velocity of 23 m/s. It is then subject to a constant acceleration of 3.5 m/s2 for 12 s. How far will it have traveled during the time of its acceleration?

Answer 1

Answer:

believe that the appropriate formula is: d = ½at² + v0t

---> d = ½(3.5)(12)² + (23)(12)

Answer 2

Given:-

• Initial velocity (u) = 23m/s

• Acceleration (a) = 3.5m/s²

• Time taken (t) = 12s

To Find:-

• Distance covered by object (s).

Solution:-

By using 2nd equation of motion

→ s = ut +1/2at²

Substitute the value we get

→ s = 23×12+1/2×3.5×12²

→ s = 276 + 1/2×3.5×144

→ s = 276 + 3.5×72

→ s = 276 + 252

→ s = 528 m

∴ The distance covered by the object is 528m.