Question

An object of mass 10 kg is at rest on a smooth
horizontal (x-y) plane. Now force
F =(9tî + 4tj)N is applied on object then
velocity of object at t = 5s is
(375 i +50)ms
(37.5i+5j)ms-1
(50i+375pms-
-1
-1
-1
(5i +37.5j)ms-​

Answer 1

the velocity of the object after 5 seconds is

$11.25\hat i+5\hat j$

Explanation:

Mass of the object m = 10 kg

Force applied on the object

$\vec F=9t\hat i+4t\hat j$ N

If the acceleration vector be [tex[\vec a[/tex]

Then we know that

$\vec F=m\vec a$

$\implies \vec a=\frac{\vec F}{m}$

$\implies \vec a=\frac{9t\hat i+4t\hat j}{10}$ m/s²

$\implies \vec a=0.9t\hat i+0.4t\hat j$ m/s²

The object is initially at rest

Therefore, initial velocity $\vec u=0$

We know that

Rate of change of velocity = acceleration

$\implies \frac{d\vec v}{dt}=\vec a$

$\implies \int d\vec v=\int_0^5 \vec adt$

$\implies \vec v=\int_0^5 (0.9t\hat i+0.4t\hat j)dt$

$\implies \vec v=(\int_0^5 0.9t dt)\hat i+(\int_0^5 0.4t)dt$

$\implies \vec v=(0.9\frac{t^2}{2}\Bigr|_0^5)\hat i+(0.4\frac{t^2}{2}\Bigr|_0^5)\hat j$

$\implies \vec v=0.9\times\frac{25}{2}\hat i+0.4\times\frac{25}{2}\hat j$

$\implies \vec v=11.25\hat i+5\hat j$

Note: The options aren't matching because some data may be wrong

Hope this answer is helpful.

Know More:

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Answer 2

Answer:Ans: V= (37.5 i+5 j)m/s

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