Please answer in detail.
At Srinagar temperature was 5°C on Monday and then it dropped by 2°C on Tuesday. What was the temperature of Srinagar on Tuesday? On Wednesday it rose by 4°C. What was the temperature on this day?
Answers
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:
Complete Step-by-Step solution:
We have been given the temperature of Srinagar on Monday = −5∘C
.
It is said that the temperature dropped by 2∘C
on Tuesday.
∴
Temperature on Tuesday = Temperature of Monday - 2∘C
.
=−5∘C−2∘C=−(5+2)∘C=−7∘C
Now we have been given that the temperature rose on by 4∘C
on Wednesday.
Thus we need to find the temperature on Wednesday.
Temperature on Wednesday = Temperature on Tuesday + 4∘C
=−7∘C+4∘C=−3∘C
Thus, we got a temperature on Tuesday = −7∘C
.
Temperature on Wednesday = −3∘C
.