Question

Two particles P and Q start their motion from same point along same straight line. P moves with constant velocity 50 m/s and Q starts moving with constant acceleration 10 m/s². Separation between them after 1s will be​

Answer 1

u

1

=50m/s

u

2

=0

s

1

=u×t

a

2

=10m/s

2

s

1

=50t

s

2

=u

2

t+

2

1

a

2

t

2

s

2

=0+

2

1

×10×t

2

=5t

2

s

1

−s

2

=125⇒50t−5t

2

=125

⇒t

2

−10t+25=0⇒t=5

Answer 2

Given,

• Two particles, P and Q, start their motion from the same point.

• P moves with a constant velocity of 50 m/s.

• Q starts moving with a constant acceleration of 10 m/s²

To find,

The separation between them after 1s

Solution,

Distance travelled in 1 second is to be calculated. We can use the formula,

            $s = ut + \frac{1}{2} at^{2}$

For particle P, there is no acceleration, $a = 0.$

$s_{P} = 50m/s*1s = 50m$

For particle Q, the initial velocity is 0, and acceleration is = $10m/s^{2}$.

$s_{Q} = \frac{1}{2}*10*1^{2}m\\s_{Q} = 5m$

Both the particles start from the same point. The separation between them is

$s_{P} - s_{Q} = 50m - 5m = 45m$

Therefore, the separation between them after 1s will be $45m.$