1
JEE Main 2021 (Online) 24th February Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language
If the velocity-time graph has the shape AMB, what would be the shape of the corresponding acceleration-time graph?

JEE Main 2021 (Online) 24th February Morning Shift Physics - Motion Question 123 English
A
JEE Main 2021 (Online) 24th February Morning Shift Physics - Motion Question 123 English Option 1
B
JEE Main 2021 (Online) 24th February Morning Shift Physics - Motion Question 123 English Option 2
C
JEE Main 2021 (Online) 24th February Morning Shift Physics - Motion Question 123 English Option 3
D
JEE Main 2021 (Online) 24th February Morning Shift Physics - Motion Question 123 English Option 4
2
JEE Main 2020 (Online) 6th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
When a car is at rest, its driver sees rain drops falling on it vertically. When driving the car with speed v, he sees that rain drops are coming at an angle 60° from the horizontal. On further increasing the speed of the car to (1 + $$\beta $$)v, this angle changes to 45o. The value of $$\beta $$ is close to :
A
0.50
B
0.73
C
0.37
D
0.41
3
JEE Main 2020 (Online) 5th September Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
The velocity (v) and time (t) graph of a body in a straight line motion is shown in the figure. The point S is at 4.333 seconds. The total distance covered by the body in 6 s is : JEE Main 2020 (Online) 5th September Evening Slot Physics - Motion Question 125 English
A
12 m
B
11 m
C
$${{49} \over 4}$$ m
D
$${{37} \over 3}$$ m
4
JEE Main 2020 (Online) 5th September Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet is dropped from the helicopter when it is at a height h. The time taken by the packet to reach the ground is close to :
[g is the acceleration due to gravity]
A
t = 3.4$$\sqrt {\left( {{h \over g}} \right)} $$
B
t = 1.8$$\sqrt {\left( {{h \over g}} \right)} $$
C
t = $$\sqrt {{{2h} \over {3g}}} $$
D
t = $${2 \over 3}\sqrt {\left( {{h \over g}} \right)} $$
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