Question

A particle initially at rest starts moving
along x-axis. Its acceleration varies with
time as a = 60 m/s2. If it starts from x =
10 m. Then the position of particle at t = 2
sis​

Answer 1

Given :

▪ Initial velocity = zero

▪ Acceleration = 60m/s²

▪ Initial position x(0) = 10m

▪ Time interval = 2s

To Find :

▪ Final position of particle after the given interval of time.

Concept :

☞ Since, acceleration has said to be constant throughout the motion, we can easily apply equation of kinematics to solve this question.

❇ Second equation of kinematics :

»» x(t) = x(0) + ut + 1/2(at²)

Where,

x(t) denotes position of body at t time

x(0) denotes initial position

u denotes initial velocity

t denotes time

a denotes acceleration

Calculation :

→ x(t) = x(0) + ut + 1/2(at²)

→ x(2) = 10 + (0×2) + 1/2[60×(2)²]

→ x(2) = 10 + 0 + (240/2)

→ x(2) = 10 + 120

→ x(2) = 130m

Answer 2

Answer:

YOUR QUESTION :

A particle initially at rest starts moving

along x-axis. Its acceleration varies with

time as a = 60 m/s2. If it starts from x =

10 m. Then the position of particle at t = 2sec

SOLUTION :

Given :

Initial velocity = zero.

Acceleration =

$60 {ms}^{2}$

Initial position = x(0) = 10m

Time interval = 2s.

To find :

Final position of particle after the given interval of time

Concept :

✔ Second equation of kinematics :

$x(t) = x(0) + ut + \frac{1}{2} {at}^{2}$

• x(t) denotes position of body at time
• x(0) denotes initial position

Calculation :

$x(t) = x(0) + ut + \frac{1}{2} {at}^{2} \\ = > x(2) = 10 + (0 \times 2) + \frac{1}{2} \: 60 \times {(2)}^{2} \\ = > x(2) = 10 + 0 + \frac{240}{2} \\ = > x(2) = 10 + 120 \\ = > x(2) = 130$

Answer : 130 m.