A car is stopped at a traffic light. It then travels along a straight road such
that its distance from the light is given by:
š„(š”) = 2.4š”
2 ā 0.120š”
3
x=2.4t2ā0.120t+3x=2.4t^2-0.120t+3x=2.4t2ā0.120t+3
Differentiate with respect to time
v=dxdtv=\frac{dx}{dt}v=dtdxā
v=d(2.4t2ā.120t+3)dt=4.8tā0.120v=\frac{d(2.4t^2-.120t+3)}{dt}=4.8t-0.120v=dtd(2.4t2ā.120t+3)ā=4.8tā0.120
v=0v=0v=0
4.8tā0.120=04.8t-0.120=04.8tā0.120=0
t=0.1204.8=0.025sect=\frac{0.120}{4.8}=0.025sect=4.80.120ā=0.025sec
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