Previous year questions: Motion in a Straight Line

Ten questions follow from the official JEE Main 2026 papers and the NCERT Exemplar. Each heading gives the paper and the marks. Attempt the question on paper before reading the answer, and write the sign convention at the top of every solution.

Q1 (JEE Main 2026, 4 Apr Shift 2, 4 marks)

Two cars A and B move in the same direction along a straight line with speeds 100 km h^-1 and 80 km h^-1, and car A is ahead of car B. A person in car B throws a stone with speed v that hits car A with a speed of 5 m s^-1. The value of v in km h^-1 is

Answer: Convert 5 m s^-1 to 18 km h^-1. In the ground frame the stone leaves B with speed 80 + v when thrown forward. Car A moves at 100 km h^-1, so the speed of the stone relative to A is (80 + v) - 100. Set that equal to 18: v - 20 = 18, so v = 38 km h^-1. The option 38 is correct.

Hinglish: 5 m/s ko 18 km/h likho. Ground frame mein stone ki speed 80 + v hai. Car A ki speed 100 km/h hai, isliye stone ki speed A ke relative (80 + v) - 100. Isko 18 ke barabar rakho: v - 20 = 18, matlab v = 38 km/h. Option 38 sahi hai.

Q2 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

The velocity-time graph of a particle over 40 s rises uniformly from zero to +5 m s^-1 in the first 10 s, falls uniformly to -5 m s^-1 by t = 30 s, and returns uniformly to zero at t = 40 s. The total distance travelled and the average velocity over the interval are

Answer: Find each area under the graph. From 0 to 10 s the area is (1/2)(10)(5) = 25 m. From 10 to 20 s the velocity falls to zero and the area is again 25 m. From 20 to 30 s the velocity is negative and the area is -25 m, and from 30 to 40 s it returns to zero with area -25 m. The total displacement is 25 + 25 - 25 - 25 = 0, so the average velocity is zero. The total distance is 25 + 25 + 25 + 25 = 100 m. The answer is 100 m and zero.

Hinglish: Graph ke areas nikaalo: 0 se 10 s tak 25 m, 10 se 20 s tak 25 m, 20 se 30 s tak -25 m aur 30 se 40 s tak -25 m. Net displacement zero hai, isliye average velocity zero. Total distance 100 m. Answer 100 m aur zero.

Q3 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

A ball is dropped from rest from a height of 18 m above the ground. The height above the ground at which the magnitude of its velocity equals the magnitude of the acceleration due to gravity, in the same set of units, is (take g = 10 m s^-2)

Answer: With g = 10 m s^-2 the magnitude of the acceleration is 10. The speed equals 10 m s^-1 when gt = 10, that is at t = 1 s. The distance fallen in 1 s is (1/2)(10)(1)^2 = 5 m. The height above the ground is 18 - 5 = 13 m. A second route is v^2 = 2gs, which gives 100 = 20s and again s = 5 m.

Hinglish: g = 10 m/s^2 hai, isliye velocity ka magnitude 10 m/s tab hoga jab gt = 10, matlab t = 1 s. 1 s mein girne ki distance 1/2 x 10 x 1 = 5 m. Height above ground 18 - 5 = 13 m. v^2 = 2gs se bhi 100 = 20s aur s = 5 m milta hai.

Q4 (JEE Main 2026, 8 Apr Shift 2, 4 marks)

A gas balloon rises with a constant velocity of 10 m s^-1. When it is at a height of 75 m a stone is dropped from it, and the balloon keeps moving up at the same velocity. The height of the balloon when the stone hits the ground is (take g = 10 m s^-2)

Answer: The stone leaves the balloon with an upward velocity of 10 m s^-1, because the balloon was rising. Take upward as positive and the release point as zero: the ground is at -75 m. Use x = ut - (1/2)gt^2, so -75 = 10t - 5t^2. This gives t^2 - 2t - 15 = 0, whose positive root is t = 5 s. In those 5 s the balloon rises another 10 x 5 = 50 m, so its height is 75 + 50 = 125 m. The answer is 125 m.

Hinglish: Stone balloon se nikala to uski initial velocity 10 m/s upar hai. Upward positive lo, release point zero, ground -75 m: -75 = 10t - 5t^2, matlab t^2 - 2t - 15 = 0 aur t = 5 s. In 5 s mein balloon 10 x 5 = 50 m aur upar gaya, isliye height 75 + 50 = 125 m.

Q5 (JEE Main 2026, 8 Apr Shift 2, 4 marks)

Two masses of 3.4 kg and 2.5 kg are accelerated from initial speeds of 5 m s^-1 and 12 m s^-1. The distances covered by them in the 5th second are 104 m and 129 m. The ratio of their momenta after 10 s is x/8. The value of x is

Answer: The distance in the nth second is u + a(2n - 1)/2. For n = 5, the acceleration of the first mass follows from 5 + a(9/2) = 104, so a = 22 m s^-2. For the second, 12 + b(9/2) = 129, so b = 26 m s^-2. After 10 s the speeds are 5 + 220 = 225 m s^-1 and 12 + 260 = 272 m s^-1. The momenta are 3.4 x 225 = 765 and 2.5 x 272 = 680, and their ratio is 765/680 = 9/8. Therefore x = 9.

Hinglish: nth second ka distance u + a(2n-1)/2. n = 5 ke liye pehle mass ka acceleration 5 + 4.5a = 104 se a = 22 m/s^2, aur doosre ka 12 + 4.5b = 129 se b = 26 m/s^2. 10 s baad speeds 225 aur 272 m/s. Momenta 765 aur 680, ratio 9/8, isliye x = 9.

Q6 (JEE Main 2026, 2 Apr Shift 2, 4 marks)

A spherical ball of mass 2 kg falls from a height of 10 m and is brought to rest after penetrating 10 cm into sand. The average force exerted by the sand on the ball is (take g = 10 m s^-2)

Answer: The speed just before the ball enters the sand is sqrt(2gh) = sqrt(200) = 14.14 m s^-1. Inside the sand it decelerates uniformly to rest over 0.1 m, so 0 = 200 + 2a(0.1) gives a = -1000 m s^-2, an upward acceleration of magnitude 1000. The net upward force needed is ma = 2000 N. That net force is the sand force minus the weight, so the sand force is 2000 + 20 = 2020 N. The answer 2020 N is correct.

Hinglish: Sand mein ghusne se pehle speed sqrt(200) = 14.14 m/s. Sand ke andar 0.1 m mein rest tak uniform deceleration: 0 = 200 + 0.2a se a = -1000 m/s^2. Net upward force ma = 2000 N. Sand ka force weight ko bhi support karta hai, isliye 2000 + 20 = 2020 N.

Q7 (NCERT Exemplar, 1 mark)

A lift is coming from the 8th floor and is just about to reach the 4th floor. Taking the ground floor as the origin and the positive direction upwards, which of the following is correct? (a) x < 0, v < 0, a > 0 (b) x > 0, v < 0, a < 0 (c) x > 0, v < 0, a > 0 (d) x > 0, v > 0, a < 0

Answer: The lift is above the ground floor, so x is positive. It is coming down, so v is negative. It is about to stop at the 4th floor, so its downward speed is decreasing, which means the acceleration points upward and is positive. The correct choice is (c).

Hinglish: Lift ground floor se upar hai, isliye x positive. Neeche aa rahi hai, isliye v negative. 4th floor par rukne wali hai, matlab speed kam ho rahi hai, isliye acceleration upar ki taraf aur positive. Answer (c).

Q8 (NCERT Exemplar, 1 mark)

A vehicle travels half the distance L with speed V1 and the other half with speed V2. Its average speed is

Answer: Each half covers a distance L/2. The times are (L/2)/V1 and (L/2)/V2, so the total time is (L/2)(1/V1 + 1/V2). The average speed is the total distance L divided by this time, which simplifies to 2V1V2/(V1 + V2). This is the harmonic mean of the two speeds and is always smaller than their arithmetic mean. The answer is the option carrying this expression.

Hinglish: Har half distance L/2 hai, isliye times (L/2)/V1 aur (L/2)/V2. Total time (L/2)(1/V1 + 1/V2), aur average speed L / total time = 2V1V2/(V1+V2). Yeh harmonic mean hai, arithmetic mean se chhota hota hai.

Q9 (NCERT Exemplar, 1 mark)

The displacement of a particle is x = (t - 2)^2, where x is in metre and t in second. The distance covered by the particle in the first 4 seconds is

Answer: The distance travelled is not the same as the change in position because the particle turns around. The position is 4 m at t = 0, zero at t = 2 and 4 m at t = 4, so the particle moves 4 m toward the origin and then 4 m back out. The distance covered is 4 + 4 = 8 m, while the displacement over the same interval is zero.

Hinglish: Distance aur displacement yahan same nahi hain kyunki particle turn karta hai. t = 0 par x = 4 m, t = 2 par x = 0 aur t = 4 par x = 4 m. Particle 4 m origin ki taraf gaya aur phir 4 m wapas, isliye distance 8 m. Same interval ka displacement zero hai.

Q10 (NCERT Exemplar, 3 marks)

A bird flies to and fro between two cars moving toward each other on a straight road. One car moves at 18 km h^-1 and the other at 27 km h^-1. The bird starts from the first car toward the other at 36 km h^-1, when the cars are 36 km apart. Find the total distance covered by the bird and its total displacement.

Answer: The cars approach each other at 18 + 27 = 45 km h^-1, so the time before they meet is 36/45 = 0.8 h. The bird flies at 36 km h^-1 for that whole time, so the distance is 36 x 0.8 = 28.8 km. The bird starts at one car and ends at the meeting point of the cars, so its displacement is the displacement of that starting car, that is 18 x 0.8 = 14.4 km toward the other car. The distance and the displacement are different because the bird reverses direction many times.

Hinglish: Cars 45 km/h ki relative speed se paas aa rahi hain, isliye milne ka time 36/45 = 0.8 h. Bird 36 km/h se itni der tak udta hai, distance 36 x 0.8 = 28.8 km. Bird apni starting car se cars ke meeting point tak pahunchta hai, isliye displacement 18 x 0.8 = 14.4 km. Beech mein bird kai baar direction badalta hai, isliye distance aur displacement alag hain.