Question

The acceleration (a) of a particular car from rest is given by the expression a = 3t( t^2 +4)^(1/2), where a is in km/h^2 and t in seconds.

(a) find the velocity at any given time (t) in units of km/h.

(b) find the time, to the nearest hundredth of a second, when the car reaches 56km/h

Answer 1

its phy qus rite then why ar u asking here

Answer 2

its differential equations

Answer 3

a) a(t) is given. Integrate it we will get v(t).
b) We know a(t), v(t) = 56km/h. At start initial velocity u(t) = 0. We need time taken
a = v-u/ t formula to find t

Answer 4

its maths qus

Answer 5

a=d2y/dx2=3t(t^2+4)^(1/2)
v=dy/dx=integral of 3t(t^2+4)^(1/2), to integrate this function use substitution method so let u=(t^2+4) therefore du=2tdt and du/2=tdt then substitute back in integral so...
integral of 3t(t^2+4)^(1/2) = 1/2 * integral of u^(1/2)du which if u integrate u get
v=(3/2)(2/3)u^(3/2)+V0, and convert the u back to t because u=t^2+4 so v=u^(3/2)+V0 becomes v=(t^2+4)^(3/2)

Answer 6

to check if u did it right just take the derivative of v=(t^2+4)^(3/2) which should give u acceleration and if u do that then you get a=3t(t^2+4)^(1/2), so it confirms v=(t^2+4)^(3/2) is the correct answer..please reply back if you have question

Answer 7

part b is you let v=(t^2+4)^(3/2)=56 and solve for t, doing that will give you 3.26 seconds for the answer

Answer 8

just integrate de equation