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find the instanteneous velocity in each of the following movements when t=5s:

I) $x=t^2+5t+2$

II) $x=t/2+3$

III) $x = t^2/4+2$

IV) $x =t^2+3t/4$

BlindxD Oct 1, 2015

Alan · Answer 1 · 2015-10-01T11:38:15

Velocity is rate of change of distance, so, if x is distance, velocity,v is given by:

l. 2t + 5

ll. 1/2

lll. t/2

lV. 2t + 3/4

Just plug in t = 5 (where approriate) to get the velocities.

BlindxD · Answer 2 · 2015-10-01T12:52:52

could you show the entire process? i also forgot to mention that lim is deltaT→0

Alan · Answer 3 · 2015-10-01T04:37:27

I'll do the first one for you:

x = t^2 + 5t + 2

v = Limit_dt→0 (x_t+dt - x_t)/dt I'll use dt instead of writing out deltat all the time.

so

Put t+dt into the expression for x to get the x_t+dt part

v = Limit_dt→0 ([t+dt]^2 + 5[t+dt] + 2 - t^2 - 5t - 2])/dt

Expand the terms in brackets

v = Limit_dt→0 (t^2 + 2tdt + dt^2 + 5t + 5dt + 2 - t^2 - 5t - 2])/dt

Subtract terms where appropriate

v = Limit_dt→0 (2tdt + dt^2 + 5dt])/dt

Divide numerator by dt

v = Limit_dt→0 (2t + dt + 5]

Let dt go to zero and you are left with:

v = 2t + 5

See if you can apply a similar approach with the others.