JEE Main Motion in Straight Line PYQ

11. (JEE Main 2023 (Online) 1st February Morning Shift )

An object moves with speed $v_{1}, v_{2}$ and $v_{3}$ along a line segment AB, BC and CD respectively as shown in figure. Where AB = BC and AD = 3AB, then average speed of the object will be:

JEE Main 2023 (Online) 1st February Morning Shift Physics - Motion Question 30 English

A. $\frac{v_{1} v_{2} v_{3}}{3 (v_{1} v_{2} + v_{2} v_{3} + v_{3} v_{1})}$

B. $\frac{(v_{1} + v_{2} + v_{3})}{3}$

C. $\frac{(v_{1} + v_{2} + v_{3})}{3 v_{1} v_{2} v_{3}}$

D. $\frac{3 v_{1} v_{2} v_{3}}{(v_{1} v_{2} + v_{2} v_{3} + v_{3} v_{1})}$

Correct Option is (D)

$A B = B C = C D$

$\begin{aligned} \Rightarrow Average speed & = \frac{Distance}{Time} \\ = \frac{A D}{\frac{A B}{V_{1}} + \frac{A B}{V_{2}} + \frac{A B}{V_{3}}} \\ = \frac{3 V_{1} V_{2} V_{3}}{V_{1} V_{2} + V_{2} V_{3} + V_{1} V_{3}} \end{aligned}$

12. (JEE Main 2023 (Online) 30th January Evening Shift )

A vehicle travels $4 km$ with speed of $3 km / h$ and another $4 km$ with speed of $5 km / h$ , then its average speed is

A. $3.75 km / h$

B. $4.25 km / h$

C. $3.50 km / h$

D. $4.00 km / h$

Correct Option is (A)

Average speed

$= \frac{Total dis \tan ce}{Total time}$

$= \frac{8}{\frac{4}{3} + \frac{4}{5}}$

$= 3.75$ km/h

13. (JEE Main 2023 (Online) 25th January Evening Shift )

The distance travelled by a particle is related to time t as $x = 4 t^{2}$ . The velocity of the particle at t=5s is :-

A. $25 {ms}^{- 1}$

B. $20 {ms}^{- 1}$

C. $8 {ms}^{- 1}$

D. $40 {ms}^{- 1}$

Correct Option is (D)

$\begin{aligned} x = 4 t^{2} \\ v = \frac{d x}{d t} = 8 t \end{aligned}$

At $t = 5 \sec$

$v = 8 \times 5 = 40 m / s$

14. (JEE Main 2023 (Online) 25th January Morning Shift )

A car travels a distance of ' $x$ ' with speed $v_{1}$ and then same distance ' $x$ ' with speed $v_{2}$ in the same direction. The average speed of the car is :

A. $\frac{v_{1} v_{2}}{2 (v_{1} + v_{2})}$

B. $\frac{2 v_{1} v_{2}}{v_{1} + v_{2}}$

C. $\frac{2 x}{v_{1} + v_{2}}$

D. $\frac{v_{1} + v_{2}}{2}$

Correct Option is (B)

$\begin{aligned} Average velocity = \frac{Total displacement}{Total time} \\ = \frac{x + x}{\frac{x}{v_{1}} + \frac{x}{v_{2}}} = \frac{2 v_{1} v_{2}}{v_{1} + v_{2}} \end{aligned}$

15.(JEE Main 2023 (Online) 30th January Morning Shift )

A horse rider covers half the distance with $5 m / s$ speed. The remaining part of the distance was travelled with speed $10 m / s$ for half the time and with speed $15 m / s$ for other half of the time. The mean speed of the rider averaged over the whole time of motion is $\frac{x}{7} m / s$ . The value of $x$ is ___________.

Correct answer is (50)

$\Rightarrow t_{1} = \frac{\frac{S}{2}}{5}$ ........ (1)

Also, $\frac{S}{2} = \frac{10 t_{2}}{2} + \frac{15 t_{2}}{2}$

$\Rightarrow t_{2} = \frac{S}{25}$ ....... (2)

$\Rightarrow$ Mean speed $= \frac{S}{t_{1} + t_{2}}$

$= \frac{S}{\frac{S}{10} + \frac{S}{25}} = \frac{250}{35}$ m/s $= \frac{50}{7}$ m/s