Answer
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Hint: We know that acceleration is how fast or slow the velocity is with respect to time.This differentiation is nothing but the slope of the graph. Hence we must take the slope of the given curves to find the correct option, among the following.
Formula used:
$a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}$
Complete answer:
We know that velocity is the rate of change of displacement with respect to time. It is mathematically denoted as $v=\dfrac{dx}{dt}$. This is nothing but the slope of the x-t graph.
Similarly, acceleration is the rate of change of velocity with respect to time. It is mathematically denoted as $a=\dfrac{dv}{dt}$. This is nothing but the slope of the v-t graph.
Given that acceleration is zero, or $a=\dfrac{dv}{dt}=0$, this happens when velocity is a constant . i.e. velocity is independent of time and its slope is $0$.
Since we know from mathematics, that $\dfrac{d}{dx}k=0$
Then the v-t graph is given as:
Then for $v$ to be a constant with respect to time we can say that the x-t graph is linear. Like $k\times x$ this is because we know that $\dfrac{d}{dx}kx=k$.
Since the x-t graph must be linear, in the given options, only C is linear and the other options are not linear i.e. specifically quadratic here.
So, the correct answer is “Option C”.
Note:
Here, the answer is discussed in the form of derivation of simplicity. However one can integrate $a=\dfrac{dv}{dt}=0$ twice, to reach the same answer. It is important to know either integration or differentiation to solve this sum. Also note that the slope of the x-t graph is nothing but the velocity and the slope of the v-t graph is the acceleration.
Formula used:
$a=\dfrac{dv}{dt}=\dfrac{d^{2}x}{dt^{2}}$
Complete answer:
We know that velocity is the rate of change of displacement with respect to time. It is mathematically denoted as $v=\dfrac{dx}{dt}$. This is nothing but the slope of the x-t graph.
Similarly, acceleration is the rate of change of velocity with respect to time. It is mathematically denoted as $a=\dfrac{dv}{dt}$. This is nothing but the slope of the v-t graph.
Given that acceleration is zero, or $a=\dfrac{dv}{dt}=0$, this happens when velocity is a constant . i.e. velocity is independent of time and its slope is $0$.
Since we know from mathematics, that $\dfrac{d}{dx}k=0$
Then the v-t graph is given as:
Then for $v$ to be a constant with respect to time we can say that the x-t graph is linear. Like $k\times x$ this is because we know that $\dfrac{d}{dx}kx=k$.
Since the x-t graph must be linear, in the given options, only C is linear and the other options are not linear i.e. specifically quadratic here.
So, the correct answer is “Option C”.
Note:
Here, the answer is discussed in the form of derivation of simplicity. However one can integrate $a=\dfrac{dv}{dt}=0$ twice, to reach the same answer. It is important to know either integration or differentiation to solve this sum. Also note that the slope of the x-t graph is nothing but the velocity and the slope of the v-t graph is the acceleration.
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