14. In one dimensional motion position of a particle from a fixed point is (say P) is + 2m at time t=0
and the particle is at P at t = 10 s. If velocity of particle is zero at t = 6 s, determine the constant
acceleration a and velocity at t = 10 s.
Answers
Answered by
2
Answer:
V
avg
(0 to 6)=
t
2
−t
1
x
2
−x
1
=
6−0
−7−(5)
=
6
−12
=−2 m/s
V
avg
(6 to 10)=
t
3
−t
2
x
3
−x
2
=
10−6
2−(−7)
=
4
9
=2.252 m/s
V
avg
(0 to 10)=
t
3
−t
1
x
3
−x
1
=
10−0
2−5
=
10
−3
=−0.3 m/s
Answered by
2
Negative, Negative, Positive (at t = 0.3 s)
Positive, Positive, Negative (at t = 1.2 s)
Negative, Positive, Positive (at t = -1.2 s)
For simple harmonic motion (SHM) of a particle, acceleration (a) is given by the relation:
a =angular frequency … (i)
t = 0.3 s
In this time interval, x is negative. Thus, the slope of the x-t plot will also be negative. Therefore, both position and velocity are negative. However, using equation (i), acceleration of the particle will be positive.
t = 1.2 s
In this time interval, x is positive. Thus, the slope of the x-t plot will also be positive. Therefore, both position and velocity are positive. However, using equation (i), acceleration of the particle comes to be negative.
t = - 1.2 s
In this time interval, x is negative. Thus, the slope of the x-t plot will also be negative. Since both x and t are negative, the velocity comes to be positive. From equation (i), it can be inferred that the acceleration of the particle will be positive.
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