Question

a body has postion x= t4+2t3+5m find its velocity at time t =1 sec​

Answer 1

ANSWER :

Given :

▪ Equation of position of a body in terms of time has provided.

▪ x = (t^4 + 2t^3 +5) m

To Find :

▪ Velocity of body at t = 1sec

Concept :

✏ We know that, On differentiating position with respect to time once, we get the velocity.

Mathematically,

☞ v = dx/dt

Calculation :

→ v = dx/dt

→ v = d(t^4+2t^3+5)/dt

→ v = 4t^3+6t^2

Putting t = 1, we get

→ v = 4(1)^3+6(1)^2

→ v = 4(1)+6(1)

→ v = 10mps

Additional information :

• Velocity is a vector quantity.
• It can be positive, negative and zero.

Answer 2

Answer:

10 m/s

Explanation:

Given an equation of a body's position in unit 'metre' such that,

$x = {t}^{4} + 2 {t}^{3} + 5$

Where, t denotes the time in second.

To find the velocity.

We know that, the velocity is the differential term of displacement or position.

So, differentiatiating the eqn, wrt time, we get

$= > \dfrac{dx}{dt} = \dfrac{d}{dt}( {t}^{4} + 2 {t}^{3} + 5) \\ \\ = >v = 4 {t}^{3} + 6 {t}^{2}$

Now, we have to find the velocity at t = 1 sec.

Substituting the value of t = 1, we get,

=> v = 4(1) + 6(1)

=> v = 10

Hence, required velocity is 10 m/s