Question

A particle with initial velocity vo - (-21 - 4j) m/s undergoes a constant acceleration a = 3 m s-2 at 37° with positive direction of x-axis. What is particle's velocity at time t = 5 second? =​

Answer 1

Answer:

Correct option is C)

Initial velocity of particle

u

= 3

i

^

+4

j

^

Acceleration

a

= 4

i

^

−3

j

^

Final velocity after 1 second

v

=

u

+

a

t

where t=1 s

⟹

v

= (3

i

^

+4

j

^

)+ (4

i

^

−3

j

^

)×1= 7

i

^

+

j

^

Speed after 1 second ∣

v

∣=

7

2

+1

=

50

m/s

Answer 2

A particle with initial velocity v₀ = (-2i - 4j) m/s undergoes a constant acceleration, a = 3 m/s² at 37° with positive direction of x - axis.

We have to find the velocity of particle at t = 5 sec.

initial velocity, v₀ = (-2i - 4j) m/s

constant acceleration, a = 3(cos37° i + sin37° j)

= 3(0.8 i + 0.6 j)

= 2.4 i + 1.8 j

let final velocity of the particle is v

time taken, t = 5 sec

now using formula,

$\vec v=\vec u+\vec at$

here,

$\vec u=v_0=(-2\hat i-4\hat j)\\\vec a=(2.4\hat i+1.8\hat j)$

$\implies\vec v=(-2\hat i-4\hat j)+(2.4\hat i+1.8\hat j)\times5\\\\=10\hat i+5\hat j$

Therefore the velocity of particle at time t = 5sec is (10i + 5j)