Question

a body starts moving in a straight line from position of rest with acceleration of 4.0 m/sec^2 find the velocity of body of after 4 second​

Answer 1

Answer:

When a body starts from rest, the acceleration of the body after 2 seconds is of its displacement.velocity change = 6.95 * 4 = 27.8 m/s . Since the initial velocity was zero, the final velocity is equal to the change of speed.

Answer 2

Answer:

Nice question!

GIVEN:

• Body is starting from rest, so initial velocity (u) = 0.
• Acceleration (a) = 4.0ms⁻².
• Time (t) = 4 seconds.

RTF:

Final velocity (v)

SOLUTION:

★Let's recall: THE THREE EQUATIONS OF MOTION ARE:

$\begin{equation} % to insert extra space between equations, you can leave space between the equation environments. But you should not leave space inside any environment. v = u + at \end{equation} \begin{equation} v^2 = u^2 + 2as \end{equation} \begin{equation} s = ut + \frac{1}{2} at^2 \end{equation} \\ where,\\ s = displacement; \\ u = initial velocity; \\ v = final velocity; \\ a = acceleration; \\ t = time of motion$$\end{center} \begin{equation} % to insert extra space between equations, you can leave space between the equation environments. But you should not leave space inside any environment. v = u + at \end{equation} \begin{equation} v^2 = u^2 + 2as \end{equation} \begin{equation} s = ut + \frac{1}{2} at^2 \end{equation} \\ where,\\ s = displacement; \\ u = initial velocity; \\ v = final velocity; \\ a = acceleration; \\ t = time of motion$$1.v=u+at\\2. s=ut+\frac{1}{2} at^{2} \\3. 2as=v^{2} -u^{2}$

We are given a, u and t, and we need to find v.

So, it is apt to use the (1)s formula.

$v=u+at$

Substitute a = 4.0ms⁻², t = 4 seconds, u = 0ms⁻¹.

$v=0+4\times 4$

$v=16 ms^{-1}$

∴FINAL VELOCITY - 16ms⁻¹.

HOPE THIS HELPS :D