Question

An object gets a variable acceleration (in m/s2) that chages with time according to the equtaion: a(t)= 5.0-1.4t 0 < t < 8.0s

a) When does the object obtain maximum velocity?

b) What is the maximum velocity

c) When is the velocity 6.5 m

d)How far has the object travelled when it obtains maxiumum velocity?

e) what distance does the object move in total from t=0 to t=8.0s ?

Answer 1

To obtain maximum velocity, let dv(t)/dt = a(t) =0

5.0-1.4t = 0

t = 5/1.4 = 3.57

d²v/dt² = da/dt = -1.4

is, the velocity will be maximum when t = 3.57 sec

b) maximum velocity:

v(t) = integral of a(t) = integral of (5.0-1.4t)dt

= 5t - 1.4t²/2 + u

= 5t - 1.4t²/2 (zero initial condition)

maximum velocity is at t = 3.57 s = 8.928 m/s

c) velocity 6.5 m/s:

v(t) = 5 = 5t - .7t²    ie, .7t² - 5t + 5 =0

solving this t = t=5.94 sec, 1.2 sec

ie, at velocity 6.5 m/s is at t=5.94 sec & 1.2 sec

d) object travelled when it obtains maxiumum velocity

s(t) = integral of v(t) = integral of (5t - .7t²)dt

= 5t²/2-.7t³/3   (zero initial condition)

s(t) at t = 3.57 sec = 21.245 m

e)object move in total from t=0 to t=8.0s:

at t = 3.57 sec object will reach maximum velocity then in next 3.57 sec (t=7.14) velocity will become zero then it will travel back

so the distance travelled in 8 sec = displacement in 7.14 sec + displacement in 7.14 to 8 sec (.86 sec)

we have s(t) =5t²/2-.7t³/3

displacement in 7.14 sec = 42.516 m

displacement in 7.14 to 8 sec (.86 sec) = 1.7 m

total distance travelled = 42.516+1.7 = 44.216 m