Why can't you use the kinematic equations to solve this problem? Is there any instruction to solve this problem by integration?
I know how to get the answer but I don't know what the formulas are
The questions goes..
You start at rest on a bike. You accelerate at 1 m/s^2 for 4 seconds and then travel at a constant velocity until you arrive at your destination which is 80 metres from the starting position. Calculate travelling time.
I know I have to integrate twice to find the distance traveled during the acceleration period but I don't know what the formula for this question will be. I have a(t) = t ? But I don't get the correct answer when I integrate..
I can do the question in my head easy but we have to use integration.
Just in case a picture helps, and if you do want or need to derive the formulae by integration from the definitions of velocity v and acceleration a in terms of displacement, s ...
... where straight lines are differentiating downwards with respect to t. Well, integrating a twice with respect to t ...
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
No, a(t) is not t. You are told that the acceleration is 1 m/s^2 for 4 seconds- a(t)= 1. What is the velocity function at each time t for those 4 seconds? What will your velocity be at the end of the 4 seconds? What will your position be at the end of the 4 seconds? What distance do you still have to go to get to a total of 80 m? How long will that take at your new velocity?
So during the period of acceleration.
v = t and will be 4 m/s at the end of the period of acceleration.
Displacement will be s = t^2/2. So displacement after 4 seconds will be 8 metres?
Leaving 72 metres left to travel at constant velocity of 4 m/s, which would take 18 seconds and then add the time taken in the acceleration period of 4 seconds to get 22 seconds all up?