An object which is initially at rest,accelerates at a rate of 4 m/s^2. Its final position is 6 m from its initial position at the end of that acceleration. For how much time did it accelerate?
Answers
Answer:
Time for which object accelerated = 1.73 sec [approx.]
Explanation:
Given:-
Initial velocity of object, u = 0 [since, it is at rest]
Acceleration of object, a = 4 m/s²
Distance covered by object, s = 6 m [since, it's final position is 6 m from its initial position at the end of acceleration]
To find:-
Time for which object accelerated, a =?
Formula required:-
Second equation of motion
s = u t + 1/2 a t²
Where,
s = distance covered
u = Initial velocity
t = time
a = acceleration
Solution:-
Using the second equation of motion
→ s = u t + 1/2 a t²
→ 6 = 0 × t + 1/2 × 4 × t²
→ 6 = 0 + 2 t²
→ t² = 6 / 2
→ t² = 3
→ t = 1.73 sec [approx.]
Therefore,Object accelerated for 1.73 seconds approximately.
Answer:
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