JEE Main 2021 (Online) 17th March Morning Shift | Motion Question 115 | Physics

A car accelerates from rest at a constant rate $$\alpha$$ for some time after which it decelerates at a constant rate $$\beta$$ to come to rest. If the total time elapsed is t seconds, the total distance travelled is :

$${{4\alpha \beta } \over {(\alpha + \beta )}}{t^2}$$

$${{2\alpha \beta } \over {(\alpha + \beta )}}{t^2}$$

$${{\alpha \beta } \over {2(\alpha + \beta )}}{t^2}$$

$${{\alpha \beta } \over {4(\alpha + \beta )}}{t^2}$$

JEE Main 2021 (Online) 16th March Evening Shift

MCQ (Single Correct Answer)

-1

A mosquito is moving with a velocity $$\overrightarrow v = 0.5{t^2}\widehat i + 3t\widehat j + 9\widehat k$$ m/s and accelerating in uniform conditions. What will be the direction of mosquito after 2 s?

$${\tan ^{ - 1}}\left( {{\sqrt {85} } \over 6}\right)$$ from y-axis

$${\tan ^{ - 1}}\left( {{5 \over 2}} \right)$$ from y-axis

$${\tan ^{ - 1}}\left( {{2 \over 3}} \right)$$ from x-axis

$${\tan ^{ - 1}}\left( {{5 \over 2}} \right)$$ from x-axis

JEE Main 2021 (Online) 16th March Morning Shift

MCQ (Single Correct Answer)

-1

The velocity-displacement graph describing the motion of bicycle is shown in the figure.

JEE Main 2021 (Online) 16th March Morning Shift Physics - Motion Question 118 English

JEE Main 2021 (Online) 16th March Morning Shift Physics - Motion Question 118 English

The acceleration-displacement graph of the bicycle's motion is best described by :

JEE Main 2021 (Online) 16th March Morning Shift Physics - Motion Question 118 English Option 1

JEE Main 2021 (Online) 16th March Morning Shift Physics - Motion Question 118 English Option 2

JEE Main 2021 (Online) 16th March Morning Shift Physics - Motion Question 118 English Option 3

JEE Main 2021 (Online) 16th March Morning Shift Physics - Motion Question 118 English Option 4

JEE Main 2021 (Online) 26th February Evening Shift

MCQ (Single Correct Answer)

-1

A scooter accelerates from rest for time t₁ at constant rate a₁ and then retards at constant rate a₂ for time t₂ and comes to rest. The correct value of $${{{t_1}} \over {{t_2}}}$$ wil be :

$${{{a_1} + {a_2}} \over {{a_2}}}$$

$${{{a_1} + {a_2}} \over {{a_1}}}$$

$${{{a_2}} \over {{a_1}}}$$

$${{{a_1}} \over {{a_2}}}$$

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Questions Asked from Motion (MCQ (Single Correct Answer))

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