A body is moving with uniform velocity of 10 ms-1. When this body just crosses another body the second one starts and moves with uniform acceleration of 5 ms-2, They meet after how much time

Question

A body is moving with uniform velocity of 10 ms-1. When this body just crosses another body the second one starts and moves with uniform acceleration of 5 ms-2, They meet after how much time (A) 3 s (B) 2 s (C) 1 s (D) 4 s

Tardigrade Admin · Accepted Answer

For first body s=ut =10 × t .....(i) For second body s = ut + (1/2) at2 hence s=0+(1/2)× 5t2......(ii) From Eqs. (i) and (ii), we get t=(20/5)=4s

Q. A body is moving with uniform velocity of $10 m s^{- 1}$ . When this body just crosses another body the second one starts and moves with uniform acceleration of $5 m s^{- 2}$ , They meet after how much time

3 s

2 s

1 s

4 s

For first body $s = u t = 10 \times t$ .....(i)
For second body s = ut + $\frac{1}{2} a t^{2}$
hence $s = 0 + \frac{1}{2} \times 5 t^{2}$ ......(ii)
From Eqs. (i) and (ii), we get
$t = \frac{20}{5} = 4 s$

Q. A body is moving with uniform velocity of 10ms−1. When this body just crosses another body the second one starts and moves with uniform acceleration of 5ms−2, They meet after how much time

3 s

2 s

1 s

4 s

For first body s=ut=10×t.....(i) For second body s = ut + 21​at2 hence s=0+21​×5t2......(ii) From Eqs. (i) and (ii), we get t=520​=4s

Q. A body is moving with uniform velocity of $10 m s^{- 1}$ . When this body just crosses another body the second one starts and moves with uniform acceleration of $5 m s^{- 2}$ , They meet after how much time

For first body $s = u t = 10 \times t$ .....(i)
For second body s = ut + $\frac{1}{2} a t^{2}$
hence $s = 0 + \frac{1}{2} \times 5 t^{2}$ ......(ii)
From Eqs. (i) and (ii), we get
$t = \frac{20}{5} = 4 s$