An object, which is initially at rest, is moving with a constant acceleration of 2.90 m/s² over a distance of 15.2 m. How long does it take for the object to move over this distance?
A. 3.23 s
B. 5.24 s
C. 2.29 s
D. 5.51 s



Answer :

Answer: A. 3.23s

Explanation:

To calculate the time it takes for the object to travel 15.2 m, we can use the following kinematics equation:

d = v₀t + 0.5at², where:

  • d is the displacement (the distance in this case)
  • v₀ is the initial velocity
  • t is the time
  • a is the acceleration

We are given:

  • d = 15.2 m
  • v₀ = 0 (since the object is initially at rest)
  • a = 2.9 m/s²

Let's plug these values into the formula:

15.2 = 0.5(2.9)t²

Solve for t:

15.2 = 1.45t²

15.2/1.45 = t²

t = 3.23 s

It takes 3.23 s for the object to move over 15.2 m.

To find the time it takes for an object to move over a given distance with constant acceleration, one can use the kinematic equation d = 1/2 at². Upon substituting the known values of distance (15.2 m) and acceleration (2.90 m/s²) and solving for time.hence the correct option is C. 2.29 s

To solve for the time it takes for an object initially at rest to move over a distance of 15.2 meters with a constant acceleration of 2.90 m/s², we can use the kinematic equation that relates distance (d), initial velocity (vi), acceleration (a), and time (t):

d = vit + 1/2 at²

Since the initial velocity is zero (vi = 0, as the object starts from rest), the equation simplifies to:

d = 1/2 at²

We are given:

Acceleration, a = 2.90 m/s²

Distance, d = 15.2 m

Plugging in the given values:

15.2 m = 1/2 (2.90 m/s²) t²

Solving for t² gives:

t² = (15.2 m) / (1.45 m/s²)

t² ≈ 10.48276 s²

Now, take the square root to find t:

t ≈ √10.48276 s²

t ≈ 3.23 s

Thus, the correct answer is A. 3.23 seconds.hence the correct option is C. 2.29 s