Question

The distance travelled by a particle moving along a straight line give by x is equal to 180 + 50 square metre find the initial velocity of the particles the velocity at the end of 4 second and the acceleration of the particle

Answer 1

Answer:

The initial velocity is 180 m/s and the acceleration is 100 m/s²

Explanation:

Given that,

The distance traveled by a particle

The position is

X = 180t+50t^2.....(I)

(I). The velocity is the rate of change of displacement.

The velocity is the first derivative of the position of the particle.

On differentiating w.r.to t equation (I)

v = \dfrac{dx}{dt}=100t+180

The velocity at t= 0

v = 180\ m/s

This is the initial velocity.

(II). The acceleration is equal to the change in velocity divided by the time taken.

Now, The velocity at t = 3 sec

v_{i} = 480\ m/s

The velocity at t = 4 sec

v_{f} = 580\ m/s

Formula of the acceleration is defined as,

a = \dfrac{v_{f}-v_{i}}{t_{f}-t_{i}}

Where, v_{f}= final velocity

v_{i}= initial velocity

t_{i}= initial time

t_{f}= final time

Put the value into the formula

a =\dfrac{580-480}{4-3}

a =\dfrac{100}{1}

a = 100\ m/s^2

Hence, The initial velocity is 180 m/s and the acceleration is 100 m/s²