Question

A body Starts with an initial velocity 'u' and moves with uniform acceleration 'a'. When the velocity has increased to 5u, the acceleration of reversed in direction but the magnitude remains the same. The body will now return to the starting point with a velocity of
A) -u
B) -6u
C) -7u
D) -9u

Please explain and answer ASAP

Answer 1

Answer:

Explanation:

The explanation already given is wrong .actually the answer is -7u.

Answer 2

Answer:

Explanation:

Concept:

The term "uniform acceleration" refers to an item travelling in a straight path with an increase in velocity at equal intervals of time.

Given:

Before shifting direction, the final velocity $v = 5u$

Acceleration speed is $a$.

Find:

Calculate the body that when will now return to its original position at a velocity.

Solution:

According to the given condition, the body will move with constant acceleration in the first case, so the displacement made by the body just before decelerating can be given as

$v^{2}-u^{2}=2 a s_{1}\\ \Rightarrow s_{1}=\frac{v^{2}-u^{2}}{2 a}=\frac{25 u^{2}-u^{2}}{2 a}$

$s_{1}=\frac{12 u^{2}}{a}$

So, in scenario II, the displacement of the body when it came to a complete stop and began travelling in the opposite direction may be calculated as follows:

$0=v^{2}-2 a s \Rightarrow 25 u^{2}=2 a s_{2}$

Now,

$\begin{aligned}&s_{2}=\frac{25 u^{2}}{2 a} \\&\therefore s_{\text {total }}=s_{1}+s_{2}=\frac{25 u^{2}}{2 a}+\frac{12 u^{2}}{a}=\frac{49 u^{2}}{2 a}\end{aligned}$

As a result, in scenario 3, when a body begins to move in the opposite direction, its final velocity will be reported as

$v_{f}^{2}=0-2 a s \Rightarrow 49 u^{2}$

$v_{f}=-7 u$

Therefore, the specified body will return to the starting place at a velocity of $7u$.

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