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(A) The acceleration-time $(a-t)$ graph shows that for $0 \le t \le 10 	ext{ s}$,acceleration $a$ increases linearly with time $t$. The equation for

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(A) The acceleration-time $(a-t)$ graph shows that for $0 \le t \le 10 	ext{ s}$,acceleration $a$ increases linearly with time $t$. The equation for this line is $a = kt$. At $t = 10 	ext{ s}$,$a = 2 	ext{ m/s}^2$,so $2 = k(10)$,which gives $k = 0.2 	ext{ s}^{-2}$. Thus,$a = 0.2t$. Since $a = dv/dt$,we have $dv = a \, dt = 0.2t \, dt$. Integrating from $t = 0$ to $t = 10 	ext{ s}$ with $v(0) = 0$,we get $v = \int_0^t 0.2t \, dt = 0.1t^2$. This is a parabolic curve (concave upwards). At $t = 10 	ext{ s}$,$v = 0.1(10)^2 = 10 	ext{ m/s}$.For $10 \le t \le 15 	ext{ s}$,the acceleration is constant at $a = 1 	ext{ m/s}^2$. Since $a = dv/dt$ is constant,the velocity-time graph will be a straight line with a positive slope of $1 	ext{ m/s}^2$. The velocity at $t = 15 	ext{ s}$ will be $v = 10 + 1(15 - 10) = 15 	ext{ m/s}$. Thus,the graph is a parabola for the first part and a straight line for the second part.

$A$ particle starts from rest and undergoes an acceleration as shown in the figure. The velocity-time graph from the figure will have which shape?

Similar Questions

What is common in the two graphs given below?

$Assertion$ : Velocity-time graph for an object in uniform motion along a straight path is a straight line parallel to the time axis.
$Reason$ : In uniform motion of an object velocity increases as the square of time elapsed.

The velocity-time graphs of a car and a scooter are shown in the figure. $(i)$ The difference between the distance travelled by the car and the scooter in $15\, s$ and $(ii)$ the time at which the car will catch up with the scooter are,respectively:

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