Question

The displacement and the increase in the velocity of a moving particle in the time interval of \( t \) to \( (t+1) s \) are 125 m and \( 50 m / s \), respectively. The distance travelled by the particle in \( (t+2)^{t h} s \) is \( \qquad \) m.

Answer 1

Given that the acceleration is constant, we start with:

\(v = u + at\)

When the velocity has increased by \(50\, m/s\), the equation becomes:

\(u + 50 = u + a \)

\(a = 50\, m/s^2\)

To find the distance traveled by the particle in (t + 2)^th second, we use:

\(S_n = u + \frac{a}{2}[2n-1]\)

For n = t + 2 (i.e., the (t + 2)^th second):

\(S_{t + 2} = u + \frac{50}{2}[2(t + 2) - 1]\)

\(S_{t + 2} = u + \frac{50}{2}[2(t + 2) - 1]\)

\(= u + 25 \times [2(t + 2)- 1]\)

\(= u + 25(2t + 4 - 1)\)

\(= u + 25 (2t +3)\)

\(= u + 50t + 75\)

\(= 100 + 75\)

\(= 175\, m\)