An object's acceleration is given by the formula

a(t) = 0.8 + 0.4e^0.2t

where a is the acceleration of the object in ms-2
and t is the time in seconds since the start of the objects motion.

if the object had a velocity of 4ms-1 after 3 seconds, how far did it travel between t = 5 and t = 6?

Question

An object's acceleration is given by the formula

a(t) = 0.8 + 0.4e^0.2t

where a is the acceleration of the object in ms-2
and t is the time in seconds since the start of the objects motion.

if the object had a velocity of 4ms-1 after 3 seconds, how far did it travel between t = 5 and t = 6?

anonymous · Answer

I got v = 0.8t + 2e^0.2t + c when i integrate the formula

primeralph · Answer

Use the boundary condition given.

primeralph · Answer

Plug in 3 for t and 4 for v. Find c.

anonymous · Answer

v(3) = 4
2.4+2e^0.6+c = 4
c = 1.6-2e^0.6
c=-2.04
is that right?

primeralph · Answer

If you did it right, then yes. Now find the average distance traveled.

anonymous · Answer

Is it alright if you double check for me?

anonymous · Answer

just in case I made a mistake that i cant see

anonymous · Answer

average distance i got was 8.38

primeralph · Answer

(V(6)-V(5))* (6-5)