If the velocity of a particle is v = At + Bt2, where A and B are constants, then the distance travelled by it between 1 s and 2 s is
- 32A + 73B
- A2 + B3
- 32A + 4B
- 3A + 7B
Answer
(1) 32A + 73B
Solution
v = dxdt = At + Bt2
x = 12 (At + Bt2) dt
= [ At22 + Bt33 ]12
= A2(4 - 1) + B3(8 - 1)
= 32A + 73B
(v does not change sign between 1 s and 2 s for positive A and B, so this displacement equals the distance.)
Answer: (1)