Question

The position (in meters) of an object moving in a straight line is given by ,s(t)=4t2+3t+14, where t is measured in seconds. What is the equation of the instantaneous velocity v(t)of the particle at time t?

Answer 1

Answer:

GIVEN:

$s = 4 {t}^{2} + 3t + 14$

TO FIND:

instantaneous velocity, v

FORMULA:

• v=ds/dt

• a=dv/dt

INSTANTANEOUS VELOCITY:

• instantaneous velocity is defined as the rate of change of displacement.

v=ds/dt

INSTANTANEOUS ACCELERATION:

• instantaneous Acceleration is defined as the rate of change of velocity

a=dv/dt

POINT TO REMEMBER:

• differentiation of displacement gives instantaneous velocity

• differentiation of velocity gives instantaneous Acceleration

• integration of acceleration gives velocity

• integration of velocity gives displacement

SOLUTION:

$v = \frac{ds}{dt} \\ \\ v = \frac{d(4 {t}^{2} + 3t + 14)}{dt} \\ \\ v = (4)(2)t + 3 + 0 \\ \\ v = 8t + 3$

ANSWER:

instantaneous velocity, v =8t+3

Answer 2

• As per the data given in the above question.
• we have to find the instantaneous Velocity v(t) of the particle at time t.

Given,

$s(t)= 4t²+3t+14$

$at \: time = t$

• Instantaneous Velocity Formula is made use of to determine the instantaneous velocity of the given body at any specific instant. It is articulated as:

$lim_∆x-0 \frac{∆x}{∆t} = \frac{dx}{dt}$

Where with respect to time t, x is the given function. The Instantaneous Velocity is articulated in m/s.

Now ,

The function is s(t)=4t²+3t+14

differentiate with respect to t,

$\frac{ds}{dt} = \frac{d}{dt} (4t²+3t+14)$

$V_inst = \frac{ds}{dt} = 4 \times \frac{d}{dt} {t}^{2} + \frac{d}{dt} 3t + \frac{d}{dt} 14$

we use the Formula ,

${x}^{n} = n {x}^{n - 1}$

$\frac{dt}{dt} = 1$

So,

$\frac{ds}{dt} = 4 \times 2t + 3 \times 1 + 0$

$\frac{ds}{dt} = 8t + 3$

Hence ,

For time t=ts, the instaneous velocity is V(t)= 8t + 3

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