An object at rest at the origin begins to mov | vedclass

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(B) For the first $4 \, s$,the object starts from rest $(u = 0)$ with uniform acceleration $a = 1 \, m/s^2$. The position $x$ as a function of time $t

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(B) For the first $4 \, s$,the object starts from rest $(u = 0)$ with uniform acceleration $a = 1 \, m/s^2$. The position $x$ as a function of time $t$ is given by $x = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(1)t^2 = \frac{t^2}{2}$. This represents a parabolic curve opening upwards starting from the origin.At $t = 4 \, s$,the velocity $v$ is $v = u + at = 0 + (1)(4) = 4 \, m/s$. The position at $t = 4 \, s$ is $x = \frac{(4)^2}{2} = 8 \, m$.For $t > 4 \, s$,the object moves with a constant velocity of $4 \, m/s$. The position $x$ as a function of time $t$ is given by $x = x_0 + v(t - t_0) = 8 + 4(t - 4) = 8 + 4t - 16 = 4t - 8$. This represents a straight line with a constant positive slope of $4 \, m/s$.The graph must show a parabolic curve for $0 \le t \le 4 \, s$ and a straight line for $t > 4 \, s$ with a continuous slope at $t = 4 \, s$. Option $(b)$ correctly depicts this behavior.

An object at rest at the origin begins to move in the $+x$-direction with a uniform acceleration of $1 \, m/s^2$ for $4 \, s$ and then it continues moving with a uniform velocity of $4 \, m/s$ in the same direction. The $x-t$ graph for the object's motion will be:

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