Question

3(t-3)=5(2t+1)
value of t​

Answer 1

Answer:

The graph in Question shows the positive acceleration.

How To Determine ?

Acceleration may be Positive and Constant or Negative and Constant for a uniformly accelerated motion.

Both the cases are discussed below;

❒ Case 1 :-

\bf \purple{ \maltese \: \: Acc. \: is \: + ve \: and \: Const.}✠Acc.is+veandConst.

⇝ Subcase 1 :-

★ When Positive Velocity is Increasing.

Positive Velocity is Increasing,

⟹ Slope of displacement - time graph is positive and increasing.

The Shape Displacement - time graph will be parabolic : x ∝ t².

a = +ve Constant

\sf \frac{dv}{dt}dtdv = +ve Constant

⟹ Slope of velocity - time graph = +ve Constant

⇝ Subcase 2 :-

★ When Negative Velocity is Decreasing.

Negative Velocity is Decreasing,

⟹ Slope of displacement - time graph is negative and decreasing.

The Shape Displacement - time graph will be parabolic : x ∝ t².

a = +ve Constant

\sf \frac{dv}{dt}dtdv = +ve Constant

⟹ Slope of velocity time - graph = +ve Constant

❒ Case 2 :-

\bf \purple{ \maltese \: \: Acc. \: is \: - ve \: and \: Const.}✠Acc.is−veandConst.

⇝ Subcase 1 :-

★ When Positive Velocity is Decreasing.

Positive Velocity is Decreasing,

⟹ Slope of displacement - time graph is positive and decreasing.

The Shape Displacement - time graph will be parabolic : x ∝ t².

a = - ve Constant

\sf \frac{dv}{dt}dtdv = - ve Constant

⟹ Slope of velocity - time graph = - ve Constant

⇝ Subcase 2 :-

★ When Negative Velocity is Increasing.

Negative Velocity is Increasing,

⟹ Slope of displacement - time graph is neagative and increasing.

The Shape Displacement - time graph will be parabolic : x ∝ t².

a = - ve Constant

\sf \frac{dv}{dt}dtdv = - ve Constant

⟹ Slope of velocity - time graph = - ve Constant.

Answer 2

Answer:

$let \: t \: be \: x \\ 3(x - 3) = 5(2x - 1) \\ 3x - 9 = 10x - 5 \\ 7x = - 4 \\ x = - 4 \div 7 \\ \\ \\ \\ hope \: this \: is \: helping \: u$