2.Motion in Straight Line
medium

A particle starts from rest. Its acceleration $(a)$ versus time $(t)$ is as shown in the figure. The maximum speed of the particle will be.....$m/s$

A$110$
B$55$
C$550$
D$660$
(IIT-2004)

Solution

(b) The area under acceleration time graph gives change in velocity. As acceleration is zero at the end of $11 \,sec$
i.e. ${v_{\max }} = $ Area of $\Delta OAB$
$ = \frac{1}{2} \times 11 \times 10 = 55\;m/s$
Standard 11
Physics

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