The acceleration (a) - time (t) graph for a b | vedclass

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(A) The change in velocity $(\Delta v)$ is equal to the area under the acceleration-time $(a-t)$ graph.Given that the body starts from rest, the initi

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) The change in velocity $(\Delta v)$ is equal to the area under the acceleration-time $(a-t)$ graph.Given that the body starts from rest, the initial velocity $(u = 0 \ m/s)$.The area under the $a-t$ graph is a right-angled triangle with base $(b = 10 \ s)$ and height $(h = 8 \ m/s^2)$.Area = $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height}$Area = $\frac{1}{2} 	imes 10 \ s 	imes 8 \ m/s^2 = 40 \ m/s$.Since $\Delta v = v_{max} - u = 40 \ m/s$ and $u = 0 \ m/s$, the maximum speed is $v_{max} = 40 \ m/s$.

Column $I$	Column $II$
$(A)$ $\frac{dv}{dt}$	$(p)$ Acceleration
$(B)$ $\frac{d\|v\|}{dt}$	$(q)$ Magnitude of acceleration
$(C)$ $\frac{dr}{dt}$	$(r)$ Velocity
$(D)$ $\left\|\frac{dr}{dt}\right\|$	$(s)$ Magnitude of velocity

The acceleration $(a)$ - time $(t)$ graph for a body starting from rest is given below. The maximum speed of the body is (in $m/s$)

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Match the following columns.
Column $I$ Column $II$
$(A)$ $\frac{dv}{dt}$ $(p)$ Acceleration
$(B)$ $\frac{d|v|}{dt}$ $(q)$ Magnitude of acceleration
$(C)$ $\frac{dr}{dt}$ $(r)$ Velocity
$(D)$ $\left|\frac{dr}{dt}\right|$ $(s)$ Magnitude of velocity

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