Question

e. A particle starts to move in a straight line from a point with the velocity of 10 metre
per second and acceleration of 2 metre per second square. Find the position and
velocity of the particle at tíme equal to 10 seconds?​

Answer 1

Displacement at t=5s is

S=ut+

2

1

at

2

=10×5+

2

1

×(−2.0)×(5)

2

=50−25=25m

i.e., after 5s, the particle will be at distance 25m from the starting point.

Velocity at t=5s is

v=u+at

or v=10+(−2)×5=0

Answer 2

Given :

Initial velocity = 10m/s

Acceleration = 2m/s²

Time interval = 10s

To Find :

Distance covered and final velocity after the given interval of time.

Solution :

Since acceleration is said to be constant, we can easily apply equation of kinematics to solve this question$.$

❖ Final velocity of particle :

➙ v = u + at

• v denotes final velocity
• u denotes initial velocity
• a denotes acceleration
• t denotes time

By substituting the given values;

➙ v = 10 + 2(10)

➙ v = 10 + 20

➙ v = 30 m/s

❖ Distance travelled by particle :

➙ d = ut + 1/2 at²

➙ d = (10 × 10) + 1/2 (2 × 10²)

➙ d = 100 + 1/2 (200)

➙ d = 100 + 100

➙ d = 200 m