A car travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the equation π₯ = 2.0π‘2 β 0.50π‘3. Calculate the distance of the car when it achieves its maximum speed in the positive x direction.Β
To be asked in question
Maximum speed achieve ix +ive x direction =?
"x=2.0t^2-0.50t^3\\rightarrow(1)"
Eqution (1) differenciate with respect to time
"v=\\frac{dx}{dt}=4t-1.50t^2\\rightarrow(2)"
Eqution (2) differenciate with respect to time
"a=\\frac{dv}{dt}=4-3t\\rightarrow(3)"
Maximum speed
a=0
"t=\\frac{4}{3}"
At "t=\\frac{4}{3}" Velocity is maximum then
Displacement
Eqution (1)put values
"x=2\\times(\\frac{4}{3})^2-0.50\\times(\\frac{4}{3})^3"
"x=\\frac{63}{27}=2.3703meter"
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