The acceleration-time graph is given. If the | vedclass

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(B) The change in velocity $\Delta v$ is equal to the area under the acceleration-time $(a-t)$ graph.The area of the triangle in the $(a-t)$ graph is

Vedclass · Accepted Answer

$15$

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(B) The change in velocity $\Delta v$ is equal to the area under the acceleration-time $(a-t)$ graph.The area of the triangle in the $(a-t)$ graph is given by:$	ext{Area} = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$$	ext{Area} = \frac{1}{2} 	imes 2\,s 	imes 10\,m/s^2 = 10\,m/s$We know that $\Delta v = v_f - v_i$,where $v_f$ is the final velocity and $v_i$ is the initial velocity.$v_f - 5\,m/s = 10\,m/s$$v_f = 15\,m/s$Thus,the velocity after $2\,s$ is $15\,m/s$.

The acceleration-time graph is given. If the initial velocity is $5\,m/s$,then the velocity after $2\,s$ is.......$m/s$.

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