The displacement x in m of four objects A, B, C and D with time t in s are given by following equations
A: x_A = 2.7t⁹ − 2.5t³
B: x_B = 6.7t² − 0.9t⁵
C: x_C = 10t − 9.2t³ + 5.7t
D: x_D = 0.8t + 1.2t² − 7.7
Which of them is/are moving with uniform acceleration?

Correct option is D

Correct answer is D
Explanation:
To determine which objects move with uniform acceleration, we examine the degree of the polynomial in the displacement equations:
For A: xA = 2.7t + 9t2 + 5t3
⇒ The highest degree is 3 (not uniform acceleration).
For B: xB = 6.7t2 + 0.9t + 5
⇒ The highest degree is 2 (uniform acceleration).
For C: xC = 10t4 + 9.2t3 + 5.7t
⇒ The highest degree is 4 (not uniform acceleration).
For D: xD = 0.8t + 1.2t2 + 7.7
⇒ The highest degree is 2 (uniform acceleration).
Therefore, the objects moving with uniform acceleration are B and D.