document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1197004>Question 1197004</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A car and a truck leave the same place and travel in the same direction along a straight road. The car starting from rest speeds up to 24 kph with a constant acceleration of 1/6 * m / (s ^ 2) and then runs at this speed. The truck leaves 40 seconds after the car with a uniform acceleration of 1/3 * m / (s ^ 2) from rest to attain a speed of 48 kph and then travels at this speed. How soon after the car started will the truck overtake the car? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #830104 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=830104\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "t = number of seconds the car has been driving\r<BR>\n" );
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document.write( "kph = kilometers per hour<BR>\n" );
document.write( "1 km = 1000 meters<BR>\n" );
document.write( "1 min = 60 sec<BR>\n" );
document.write( "1 hour = 60 min<BR>\n" );
document.write( "1 hour = (60*60) sec<BR>\n" );
document.write( "1 hour = 3600 sec\r<BR>\n" );
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document.write( "Because the acceleration is in m/s^2, but the velocities are in kph, I'll convert 24 kph to m/s<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?24 \text{ kph} = \frac{24 \ \text{km}}{1 \ \text{hr}}\times\frac{1000 \ \text{m}}{1 \ \text{km}}\times\frac{1 \ \text{hr}}{3600 \ \text{sec}} = \frac{20}{3} \text{ meters per second}\">\r<BR>\n" );
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document.write( "We'll use a kinematics equation to find the time it takes for the car to reach a velocity of (20/3) m/s.\r<BR>\n" );
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document.write( "Vi = initial velocity = 0<BR>\n" );
document.write( "Vf = final velocity = 20/3<BR>\n" );
document.write( "a = acceleration = 1/6<BR>\n" );
document.write( "t = (Vf - Vi)/a<BR>\n" );
document.write( "t = (20/3 - 0)/(1/6)<BR>\n" );
document.write( "t = 40<BR>\n" );
document.write( "It takes 40 seconds for the car to reach 20/3 meters per second, aka 24 kph<BR>\n" );
document.write( "When the truck starts up, the car has reached its target speed of 24 kph.\r<BR>\n" );
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document.write( "Now use another kinematics equation to compute the distance traveled during this time for the car.<BR>\n" );
document.write( "d = ( (vf)^2 - (vi)^2 )/(2a)<BR>\n" );
document.write( "d = ( (20/3)^2 - (0)^2 )/(2*1/6)<BR>\n" );
document.write( "d = 400/3<BR>\n" );
document.write( "The car has traveled 400/3 meters over the course of 40 seconds when it was accelerating from 0 to 24 kph.\r<BR>\n" );
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document.write( "As the car is accelerating, we cannot use distance = rate*time since the rate aka speed is changing.\r<BR>\n" );
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document.write( "After the t = 40 second mark, the car is now traveling at a constant velocity. <BR>\n" );
document.write( "At this point we can use distance = rate*time<BR>\n" );
document.write( "The car travels a further (20/3)*(t-40) meters where t > 40\r<BR>\n" );
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document.write( "In total the car travels a distance of 400/3 + (20/3)*(t-40) meters when t > 40\r<BR>\n" );
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document.write( "--------------------------------------------------------\r<BR>\n" );
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document.write( "Now we move to the truck. We'll follow the same outline as we did with the car in the previous section.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?48 \text{ kph} = \frac{48\ \text{km}}{1 \ \text{hr}}\times\frac{1000 \ \text{m}}{1 \ \text{km}}\times\frac{1 \ \text{hr}}{3600 \ \text{sec}} = \frac{40}{3} \text{ meters per second}\"><BR>\n" );
document.write( "or note that because 48 = 2*24, i.e. the truck's final velocity is twice that of the car's final velocity, we can say:  2*(20/3) = 40/3\r<BR>\n" );
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document.write( "Calculate the time duration throughout the acceleration period<BR>\n" );
document.write( "Vi = initial velocity = 0<BR>\n" );
document.write( "Vf = final velocity = 40/3<BR>\n" );
document.write( "a = acceleration = 1/3<BR>\n" );
document.write( "t = (Vf - Vi)/a<BR>\n" );
document.write( "t = (40/3 - 0)/(1/3)<BR>\n" );
document.write( "t = 40<BR>\n" );
document.write( "The truck also needs 40 seconds to accelerate to the target velocity it's after.<BR>\n" );
document.write( "This is expected because the truck is going twice as fast, and its acceleration is twice as much<BR>\n" );
document.write( "2*(1/6) = 1/3<BR>\n" );
document.write( "Effectively the '2's cancel out and that's why we end up with t = 40 like from before.\r<BR>\n" );
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document.write( "Then calculate the distance during this acceleration period<BR>\n" );
document.write( "d = ( (Vf)^2 - (Vi)^2 )/(2a)<BR>\n" );
document.write( "d = ( (40/3)^2 - (0)^2 )/(2*1/3)<BR>\n" );
document.write( "d = 800/3\r<BR>\n" );
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document.write( "The total distance the truck travels is <BR>\n" );
document.write( "800/3 + (40/3)*(t-80) where t > 80\r<BR>\n" );
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document.write( "The t-80 refers to the idea where the truck waited 40 seconds after the car started. <BR>\n" );
document.write( "Then it takes another 40 more seconds for the truck to get to the speed of 40/3 meters per second (48 kph)<BR>\n" );
document.write( "In total, the truck has reached the timestamp of 40+40 = 80 seconds once it has reached the desired velocity.\r<BR>\n" );
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document.write( "--------------------------------------------------------\r<BR>\n" );
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document.write( "We have these two distance expressions<BR>\n" );
document.write( "Car: 400/3 + (20/3)*(t-40)  when t > 40<BR>\n" );
document.write( "Truck: 800/3 + (40/3)*(t-80)  when t > 80<BR>\n" );
document.write( "For both of them to make sense, we need t > 80, which is where the two intervals overlap.\r<BR>\n" );
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document.write( "Set the two expressions equal to one another to see when they'll meet up<BR>\n" );
document.write( "400/3 + (20/3)*(t-40) = 800/3 + (40/3)*(t-80)<BR>\n" );
document.write( "3 * [ 400/3 + (20/3)*(t-40) ] = 3 * [ 800/3 + (40/3)*(t-80) ]<BR>\n" );
document.write( "400 + 20*(t-40) = 800 + 40*(t-80)<BR>\n" );
document.write( "400 + 20t - 800 = 800 + 40t - 3200<BR>\n" );
document.write( "20t - 400 = 40t - 2400<BR>\n" );
document.write( "20t-40t = -2400+400<BR>\n" );
document.write( "-20t = -2000<BR>\n" );
document.write( "t = -2000/(-20)<BR>\n" );
document.write( "t = 100\r<BR>\n" );
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document.write( "It takes 100 seconds for the truck to meet up with the car. After this point in time the truck passes by the car.\r<BR>\n" );
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document.write( "Recall that the start point is from the car's perspective. In other words, once the car starts driving is when the clock starts. The truck will have been driving for t-40 = 100-40 = 60 seconds when the truck meets the car, due to the 40 second head start the car had.\r<BR>\n" );
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document.write( "Answer: <font color=red>100 seconds</font>\r<BR>\n" );
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document.write( "Extra info:<BR>\n" );
document.write( "100 seconds = 60 seconds + 40 seconds = 1 minute + 40 seconds<BR>\n" );
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