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? =
-
Subject:
Physics -
Author:
garrettblack -
Created:
1 year ago
Answers 2
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
-
Author:
shayleerollins
-
Rate an answer:
4
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,
[tex]\vec v=\vec u+\vec at[/tex]
here,
[tex]\vec u=v_0=(-2\hat i-4\hat j)\\\vec a=(2.4\hat i+1.8\hat j)[/tex]
[tex]\implies\vec v=(-2\hat i-4\hat j)+(2.4\hat i+1.8\hat j)\times5\\\\=10\hat i+5\hat j[/tex]
Therefore the velocity of particle at time t = 5sec is (10i + 5j)-
Author:
borismendoza
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1
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Here is the full answer:
Refer to the attachment.
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