Question

A body is thrown vertically upwards. Which one of the following graphs correctly represents the velocity vs time?

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Step 1: Given data

A body is thrown vertically upwards

Step 2: Formula used

$a=\frac{dv}{dt}[wherea=acceleration,\frac{dv}{dt}=changeinvelocityovertime]$

Step 3: Determining the velocity-time graph.

In the time "t seconds," we have a body that is propelled upwards at a velocity of "v m/s."
We must now determine the shape of the resulting velocity-time curve.

Consider a body that is projected upwards as shown in the above figure.
Its initial velocity, $v=u_y$. We can state that $v=u_y$ since the body is thrust in the $y$ direction.
We know that when a body moves above, gravity's acceleration acts downwards. The acceleration owing to gravity will be the same as the body's acceleration. $a=g$, where $a$ denotes body acceleration and $g$ denotes gravitational acceleration.
Step 4: Differentiation

We all know that the first derivative of velocity with respect to time is acceleration.

$$\begin{gathered} a=\frac{dv}{dt}=g \\ \frac{dv}{dt}=g \\ dv=g\times dt \end{gathered}$$

Step 5: Integration

Now, let us integrate both sides of the equation.${\int }_u^{v}dv={\int }_0^{g}gdt[giscons\tan t]$

By integrating we get,

$$\begin{gathered} v-u=gt \\ v=u+gt \end{gathered}$$

We know that gravitational acceleration, abbreviated as 'g,' acts downwards. As a result, $g$ is negative.
As a result, the equation $v=u+gt$ can be written as $v=u-gt$.

Step 6: Equation of a Straight Line

Let's look at the equation for a straight line.
The straight-line equation is $y=mx+c$, where $m$ is the slope and $c$ is the $y$ axis intercept.
Let us now compare the straight-line equation to the equation we obtained, $v=u-gt$.
We can deduce that the body's velocity-time curve is a straight line with slope $-g$ and $y$ intercept $u$ when projected upwards.

Hence, option D is the correct answer.