Question

Given the following speeds versus elapsed time, for a 1000 kg object accelerating horizontally with negligible friction and constant propulsive power such that applied work on the objectis power x elapsed time. Instantanous power is defined as force x velocity.

Hint: the kinetic energy at time zero is 16 kJ.

Time (s): t0=0, t1=1, t2=2, t3=3, Speed (m/s) V0= 5.66, V1= 8, V2=9.80, V3=unknown.

The acceleration at t1=1 s with V1=8 m/s is, a3= ____________ (m/s^2).

I will rate if you give me the right answer. Thank you.

Answer #1

KE initial = 16000J

and KE initial = 0.5m* Vo^2 = 16000

Vo = sqrt(32) = 5.66 m/s

the velocity equation is interms of time

V = at^2 + bt + c

t = 0, Vo = 5.66

so, V = 0 + 0 + c = 5.66, c = 5.66

t1 = 1, V1 = 8

so, 8 = a + b + 5.66

a + b = 2.34

t2 = 2, V2 = 9.8

9.8 = 4a + 2b + 5.66

2a + b = 2.07

solving them gives a = -0.27

b - 0.27 = 2.34

b = 2.61

V = -0.27t^2 + 2.61t + 5.66

acceleration a = dV/dt = -0.54t + 2.61 = -0.54t + 2.61 = 2.07 m/s2 (t = 1s)

at t= 3s, V3 = - 0.27*9 + 7.83 + 5.66 = 11.06 m/s

and t = 3s, a3 = -0.54t + 2.61 = -0.54*3 + 2.61 = 0.99 m/s2

Given the following speeds versus elapsed time, for a 1000 kg
object accelerating horizontally with constant propulsive power and
negligible friction (hint - plot KE vs time & notice units of
Joule vs kJ)
Time (s) t0 = 0, t1 = 1, t2 = 2, t3 = 3, Speed (m/s) V0 = 5.66,
V1 = 8.00, V2 = 9.80, V3 = unknown
The propulsive force at time zero is _____ (kN)

Given the following speeds versus elapsed time, for a 1000 kg
object accelerating horizontally with constant propulsive power and
negligible friction (hint - plot KE vs time & notice units of
Joule vs kJ)
Time (s) t0 = 0, t1 = 1, t2 = 2, t3 = 3, Speed (m/s) V0 = 5.66,
V1 = 8.00, V2 = 9.80, V3 = unknown
The constant propulsive power is ____ (kW). 1 kW = 1 kJ/s = 1000
J/s = 1000 Watts...

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