C2 MEI OCR maths question
Hi guys, im on the last page of the textbook and I can't seem to solve this question. Help!!
"The velocity 'v'ms^-1 of a car 't' seconds after it passes point A has gradient function
dv/dt= 2/t^0.5 for t>0
Given that the car has velocity 5ms^-1 as it passes A, find the equation for the velocity at time 't' seconds."
My working so far:
v=4t^0.5+c
stumped to find c.
Textbook says in answers that c=5 but no explanation is given.
"The velocity 'v'ms^-1 of a car 't' seconds after it passes point A has gradient function
dv/dt= 2/t^0.5 for t>0
Given that the car has velocity 5ms^-1 as it passes A, find the equation for the velocity at time 't' seconds."
My working so far:
v=4t^0.5+c
stumped to find c.
Textbook says in answers that c=5 but no explanation is given.
Original post by FrustratedHorse
Hi guys, im on the last page of the textbook and I can't seem to solve this question. Help!!
"The velocity 'v'ms^-1 of a car 't' seconds after it passes point A has gradient function
dv/dt= 2/t^0.5 for t>0
Given that the car has velocity 5ms^-1 as it passes A, find the equation for the velocity at time 't' seconds."
My working so far:
v=4t^0.5+c
stumped to find c.
Textbook says in answers that c=5 but no explanation is given.
"The velocity 'v'ms^-1 of a car 't' seconds after it passes point A has gradient function
dv/dt= 2/t^0.5 for t>0
Given that the car has velocity 5ms^-1 as it passes A, find the equation for the velocity at time 't' seconds."
My working so far:
v=4t^0.5+c
stumped to find c.
Textbook says in answers that c=5 but no explanation is given.
Check the question again for information given that will help you.
(The number pops up in the question... why?)
yup, still fully stumped-sorry!
Hi, first of all you've integrated the dv/dx to get 4t^0.5 + c correctly.
It says that t is the time AFTER passing point A, and the question says that the velocity is 5 when it is at A, which means t is 0 at that point.
If you sub t = 0 into the equation, you get 5 = 4(0)^0.5 + c...
0 to the power of anything is 0, so...
5 = 0 + c
so c = 5
hope that helps
It says that t is the time AFTER passing point A, and the question says that the velocity is 5 when it is at A, which means t is 0 at that point.
If you sub t = 0 into the equation, you get 5 = 4(0)^0.5 + c...
0 to the power of anything is 0, so...
5 = 0 + c
so c = 5
hope that helps
Thanks, it really helped. I thought you could sub any time in as long as it was above 0 but I obviously missed out that specific information. I can finish the textbook now!
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