A particle starts with an initial velocity of 10 m/s towards north. It experience a constant acceleration of 4 m/s^2 towards south. The distance travelled by the particle in 3rd second is ?
a)zero
b)1m
c)0.5m
d)0.25m
Answers
Answer:
t=2π√(L cosθg) Show that the expression of the time period T of a simple pendulum of length l given by T=2π√lg is dimensionally correct. The periodic time (T) of a simple pendulum of length (L) is given by T=2π√Lg.
Explanation:
t=2π√(L cosθg) Show that the expression of the time period T of a simple pendulum of length l given by T=2π√lg is dimensionally correct. The periodic time (T) of a simple pendulum of length (L) is given by T=2π√Lg.
Answer :
- The distance travelled by the particle in 3rd second is 0
- Option (a)
Given :
- A particle starts with an initial velocity of 10 m/s towards north. It experience a constant acceleration of 4m/s² towards south.
To find :
- The distance travelled by the particle in 3rd second is ?
Solution :
Given :
Here , object is travelling with 10m/s towards the north and acceleration is towards south then acceleration retards the body and also acceleration will be negative.
- Acceleration (a) = -4m/s²
- Object Velocity (u) = 10m/s
We know that formula :
- Distance = u + a/2 (2n - 1)
Where,
- u is velocity object
- a is acceleration
- n is distance travelled by the particle in 3rd second
This is distance travelled by body in nth second formula
And also Given , that the distance travelled by the particle in 3rd second so,
↝ Distance = u + a/2 (2n - 1)
↝ Distance = 10 + -4/2 (2(3) - 1)
↝ Distance = 10 + -2 (5)
↝ Distance = 10 - 10
↝ Distance = 0
Hence, The distance travelled by the particle in 3rd second is 0.