Questions: Motion in a Straight Line

Q1. Distinguish between distance and displacement, and between average speed and average velocity.

Answer: Distance is the total path length covered and is positive. Displacement is the change of position from the starting point to the final point, carries a sign that encodes direction, and can be zero even after a long journey. Average speed divides the path length by the time, while average velocity divides the displacement by the time. The average speed is always at least as large as the magnitude of the average velocity, and the two are equal only when the motion never reverses direction.

Hinglish: Distance poora rasta hai, scalar aur hamesha positive. Displacement starting se final position ka fark hai, sign ke saath, aur lambi journey ke baad bhi zero ho sakta hai. Average speed path length/time hai, average velocity displacement/time. Average speed kabhi average velocity ke magnitude se chhoti nahi hoti, aur dono tabhi barabar hoti hain jab direction kabhi na badle.

Q2. The position of a particle is x = 4t^2 - 2t, with x in metre and t in second. Find its velocity at t = 2 s and its acceleration.

Answer: Differentiate the position once: v = dx/dt = 8t - 2, so at t = 2 s the velocity is 14 m s^-1. Differentiate again: a = dv/dt = 8 m s^-2, which is constant. The particle is uniformly accelerated, so the kinematic equations apply to it.

Hinglish: Position ko ek baar differentiate karo: v = 8t - 2, isliye t = 2 s par velocity 14 m/s. Dobara differentiate karne par a = 8 m/s^2, jo constant hai, matlab motion uniformly accelerated hai. t = 0 par velocity -2 m/s thi, matlab particle pehle negative direction mein chala aur 8t - 2 = 0 par turn hua.

Q3. A car accelerates uniformly from 18 km h^-1 to 72 km h^-1 in 10 s. Find its acceleration and the distance covered in that time.

Answer: Convert the speeds: 18 km h^-1 is 5 m s^-1 and 72 km h^-1 is 20 m s^-1. The acceleration is (20 - 5)/10 = 1.5 m s^-2. The distance follows from the area form: x = (u + v)t/2 = (5 + 20)(10)/2 = 125 m. Using x = ut + (1/2)at^2 gives the same 50 + 75 = 125 m, which is a useful check on the algebra.

Hinglish: Speeds ko convert karo: 18 km/h = 5 m/s aur 72 km/h = 20 m/s. Acceleration (20-5)/10 = 1.5 m/s^2. Distance x = (u+v)t/2 = 25 x 10/2 = 125 m. ut + 1/2 a t^2 se bhi 50 + 75 = 125 m aata hai, isliye dono method check kar lo.

Q4. A ball is thrown vertically upward with a speed of 20 m s^-1. Find the maximum height, the time to reach it, and the total time of flight. Take g = 10 m s^-2.

Answer: At the highest point the velocity is zero, so 0 = 400 - 2(10)h gives h = 20 m. The time to the top follows from v = u - gt, and 0 = 20 - 10t gives t = 2 s. The downward journey takes the same time because the acceleration is the same g, so the total flight time is 4 s. The ball returns to the hand with speed 20 m s^-1 downward, and the displacement over the whole flight is zero although the path length is 40 m.

Hinglish: Highest point par velocity zero hoti hai, isliye 0 = 400 - 20h se h = 20 m. Upar jaane ka time 0 = 20 - 10t se t = 2 s. Neeche aane mein bhi utna hi time lagta hai, isliye total 4 s. Ball 20 m/s ki speed se wapas aati hai, poori journey ka displacement zero hai par path length 40 m.

Q5. Why can a position-time graph never be a closed loop? What does a crossing of two position-time curves mean?

Answer: A closed loop would make the body return to an earlier position at the same time, so at some instant it would have two positions at once. That is impossible for a particle, and it is why the graph must be single valued. The slope of the graph gives the velocity, so a loop would also imply two velocities at one instant. When two position-time curves cross, both bodies are at the same position at that instant, which is what overtaking means in one dimension.

Hinglish: Closed loop ka matlab hota ki body ek hi time par do jagah ho, jo possible nahi. Position-time graph ko single valued hona padta hai, warna ek instant par do velocities banengi. Do x-t curves ka crossing point matlab us waqt dono bodies same position par hain, yani ek ne doosri ko overtake kiya.

Q6. A body starting from rest travels 100 m in 5 s with uniform acceleration. Find the acceleration and the distance travelled in the next 5 s.

Answer: From x = (1/2)at^2 with u = 0, 100 = (1/2)a(25), so a = 8 m s^-2. At 5 s the velocity is 40 m s^-1. For the next 5 s the body travels 40(5) + (1/2)(8)(25) = 200 + 100 = 300 m, so the total distance in 10 s is 400 m. The second five seconds covers three times the first, which matches the ratio of distances in equal intervals for motion from rest.

Hinglish: x = 1/2 a t^2 se 100 = 12.5a, isliye a = 8 m/s^2. 5 s par velocity 40 m/s. Agle 5 s mein distance 40 x 5 + 1/2 x 8 x 25 = 200 + 100 = 300 m, matlab 10 s mein total 400 m. Rest se chalne wali body ke liye barabar intervals mein distance ka ratio badhta jaata hai.

Q7. Explain how the displacement can be found from a velocity-time graph, taking the sign of the area into account.

Answer: The area between the velocity-time curve and the time axis equals the displacement over that interval. The derivation is that a small strip of width dt and height v has area v dt, which is exactly the displacement during dt, so summing the strips integrates v with respect to t. Areas above the axis count as positive displacement and areas below count as negative displacement. If the curve crosses the axis, the two regions have opposite signs, so the net displacement is their difference even though the distance travelled is their sum.

Hinglish: v-t graph aur time axis ke beech ka area displacement deta hai, kyunki chhoti strip ka area v dt displacement hi hai. Axis ke upar ka area positive displacement aur neeche ka negative. Curve axis ko cross kare to dono regions ka sign ulta hota hai, isliye net displacement difference se aur distance sum se milta hai. Isi wajah se displacement zero aur path length bada ho sakta hai.

Q8. A stone is dropped from a cliff and reaches the ground in 3 s. Find the height of the cliff and the speed on impact. Take g = 10 m s^-2.

Answer: The stone starts from rest, so h = (1/2)gt^2 = (1/2)(10)(9) = 45 m. The impact speed is gt = 30 m s^-1, which also follows from v^2 = 2gh = 2(10)(45) = 900. Both routes agree. The motion is uniformly accelerated, so the average velocity over the fall is 15 m s^-1 and the height is that average multiplied by the time, which is another way to see 45 m.

Hinglish: Stone rest se girta hai, isliye h = 1/2 g t^2 = 1/2 x 10 x 9 = 45 m. Impact speed gt = 30 m/s, aur v^2 = 2gh = 900 se bhi wahi. Motion uniformly accelerated hai, average velocity 15 m/s hai, aur 15 x 3 = 45 m se height cross check ho jaati hai.

Q9. Two trains of length 100 m each move in opposite directions on parallel tracks at 20 m s^-1 and 15 m s^-1. How long do they take to cross each other?

Answer: In opposite directions the relative speed is the sum, 35 m s^-1. The distance to be covered for the trains to clear each other is the sum of their lengths, 200 m. The crossing time is therefore 200/35 = 5.7 s, approximately. If they were moving in the same direction the relative speed would be the difference, 5 m s^-1, and the crossing would take much longer at the same total length.

Hinglish: Opposite direction mein relative speed sum hoti hai, 20 + 15 = 35 m/s. Ek doosre ko cross karne ke liye distance dono lengths ka sum, 200 m. Time = 200/35 = 5.7 s. Same direction mein relative speed difference 5 m/s hoti, isliye crossing bahut lambi lagti.

Q10. A body has zero velocity and non-zero acceleration at an instant. Is this possible? Give an example and explain the physical meaning.

Answer: It is possible, and a ball at the highest point of a vertical throw is the standard example. The velocity is instantaneously zero, but the acceleration is g downward, so the very next instant the ball has a downward velocity. Zero velocity at one instant says nothing about the rate of change of velocity, which is the acceleration.

Hinglish: Possible hai, aur vertical throw mein ball ka highest point iska standard example hai. Us instant velocity zero hai par acceleration g neeche ki taraf, isliye agle hi instant ball ki velocity neeche ki taraf ho jaati hai. Ek instant par velocity zero hona acceleration ke baare mein kuch nahi batata. Aage peeche hilne wali body ke har turning point par velocity zero aur acceleration maximum hota hai.

Quiz below: five questions on displacement against distance, graph slopes, kinematic equations, free fall and relative velocity. Attempt them before looking at the formula sheet again.