Question

Starting from rest the acceleration of a particle a =2(t-1) . The velocity of the particle after t=5 second is

Answer 1

Answer:

15 m/s

Explanation:

Body is starting from rest

so, u=0

a=2(t-1)

i.e. a= 2t-2=dv/dt

by integrating both side

v-u=2t^2/2 - 2t

v= t^2 - 2t

= (5)^2 - 2*5

= 25 - 10

= 15

so, velocity after 5 second is 15 m/s

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

a = 2(t - 1)

a = 2t - 2 m/s²

u = 0 m/s.

v =?.

t = 5 seconds.

$\huge{\bold{\underline{Explanation:-}}}$

As we need to find velocity from acceleration we need to use integration concept.

As given,

$\large{\bold{a = 2t - 2}}$

Now,

$\large{a = \frac{dv}{dt}}$

$\large{dv = a.dt}$

Integrating

$\large{ \implies \displaystyle \int dv = \displaystyle \int a.dt}$

Applying limits.

$\large{ \implies \displaystyle \int ^v_0 dv = \displaystyle \int^5_0 a.dt}$

Substituting the value of acceleration.

$\large{ \implies \displaystyle \int ^v_0 dv = \displaystyle \int^5_0 (2t - 2)dt}$

Simplifying

$\large{ \implies v = \left|\frac{2t^2}{2} - 2t \right|^5_0}$

Now,

$\large{ \implies v = \left|\frac{ \cancel{2}t^2}{ \cancel{2}} - 2t \right|^5_0}$

$\large{ \implies v = \left|{t^2} - 2t \right|^5_0}$

Applying limits

$\large{ \implies v = 25 - 10}$

$\huge{\boxed{\boxed{v = 15 \: m/s}}}$

So, the velocity of the particle is 15 m/s.