Question

The acceleration of a body is defined as the ratio of the change in velocity of the body to the time taken to change this velocity. The displacement of a particle starting from rest (@t=0) is given by x=3t2-t3 where x is in m and t is in sec. QUESTION 1 The velocity of the body @t=1sec is?

3m/s

6m/s

9m/s

0

Answer 1

Answer:

The velocity of the body at t = 1 sec is 3 m/s

Explanation:

Given :

The displacement of a particle starting from rest (at t = 0) is given by x = 3t² - t³ where x is in m and t is in sec.

To find :

the velocity of the body at t = 1 sec

Solution :

The velocity of a body is the ratio of the change in displacement to the time taken.

 Differentiation of displacement gives us the velocity.

 $\longrightarrow \boxed{\tt v=\dfrac{dx}{dt}}$

So, let's differentiate the displacement

$\tt v=\dfrac{dx}{dt} \\\\ \tt v=\dfrac{d}{dt}(3t^2 - t^3) \\\\ \tt v=6t-3t^2$

The velocity of the body is (6t - 2t²)

At t = 1 sec.

velocity = 6(1) - 3(1)²

velocity = 6 - 3

velocity = 3 m/s

∴ The velocity of the body at t = 1 sec is 3 m/s

Answer 2

Answer:

$\orange{ \boxed{\sf \: velocity \: \: v \: = 3 \: m {s}^{ - 1} }}$

Explanation:

Given,

The displacement of a particle starting from rest is given by

$\sf \: x = 3 {t}^{2} - {t}^{3} \: \: \: \: \: - - - (1)$

To find :

The velocity of the body at t = 1 sec .

Solution :

To find velocity, we have to differentiate eqn.(1) respect to ' t '.

Now,

$\: \: \: \: \: \: \sf \: x = 3 {t}^{2} - {t}^{3} \\ \\ \implies \: \sf \frac{d \: x}{dt} = \frac{d \: (3 {t}^{2} - {t}^{3}) }{dt} \\ \\ \implies \sf \: v = 6t - 3 {t}^{2} \\ \\ \: \sf \therefore \: velocity \: at \: \: \: t \: = 1 \: sec \\ \\ \sf \: v_{(t = 1)} = 6 \times 1 - 3 \times {1}^{2} \\ \: \: \: \: \: \: \: \: \: \: \: \: \sf= 3 \: m {s}^{ - 1}$