We read your photo as:

An object takes off from rest and accelerates at 2 m/s² to a final velocity of 10 m/s. What distance does it cover?

Physics

ScanSolve AI Answer

Answer

A 25 m\text{A } \textbf{25 m}

Step-by-step solution

4 steps
  1. Identify the given values from the problem.

    The object takes off from rest, so the initial velocity (uu) is 0 m/s0 \text{ m/s}.
    The acceleration (aa) is 2 m/s22 \text{ m/s}^2.
    The final velocity (vv) is 10 m/s10 \text{ m/s}.
    We need to find the distance covered (ss).

  2. Select the appropriate kinematic equation.

    The equation that relates initial velocity (uu), final velocity (vv), acceleration (aa), and distance (ss) is:

    v2=u2+2asv^2 = u^2 + 2as
  3. Substitute the given values into the equation.

    (10 m/s)2=(0 m/s)2+2×(2 m/s2)×s(10 \text{ m/s})^2 = (0 \text{ m/s})^2 + 2 \times (2 \text{ m/s}^2) \times s
  4. Solve for the distance (ss).

    100 m2/s2=0+4 m/s2×s100 \text{ m}^2/\text{s}^2 = 0 + 4 \text{ m/s}^2 \times s 100 m2/s2=4 m/s2×s100 \text{ m}^2/\text{s}^2 = 4 \text{ m/s}^2 \times s s=100 m2/s24 m/s2s = \frac{100 \text{ m}^2/\text{s}^2}{4 \text{ m/s}^2} s=25 ms = 25 \text{ m}

    The distance covered is 25 m25 \text{ m}. This corresponds to option A.

    The final answer is A 25 m\text{A } \textbf{25 m}.

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp