Exams › JEE Main › Physics › Motion in a Straight Line
161 questions with worked solutions.
Answer: 100 second
Closing speed = 10 - 9 = 1 m/s and the gap is 100 m, so time = 100/1 = 100 s.
Answer: 17.2 s
The total time taken includes the time to accelerate to 108 km/h, the time spent at constant speed, and the time to decelerate over 45 m. Calculating each segment shows that the total time sums up to 17.2 seconds, confirming the correct option.
Answer: v0 / √((2v0²kt)+1)
dv/dt = -k v^3 gives v^-3 dv = -k dt. Integrating from v0 to v: (1/v^2 - 1/v0^2)/2 = k t, so 1/v^2 = 1/v0^2 + 2kt. Hence v = v0/sqrt(1 + 2 v0^2 k t).
Answer: 78.4 m
The two balls meet when they have traveled the same vertical distance. The ball dropped from the tower falls 78.4 m, leaving 21.6 m above the ground, while the ball thrown upward reaches the same height after traveling 21.6 m upward, confirming that they meet 78.4 m below the top of the tower.
Answer: v0 + g/2 + f/3
x = integral_0^1 (v0 + g*t + f*t^2) dt = v0 + g/2 + f/3. The t^2 term integrates to f/3, not f/2.
Answer: 24 m
With the same maximum deceleration, stopping distance d is proportional to v^2 (d = v^2/2a). Doubling the speed (50 -> 100 km/h) multiplies the distance by 4: 4 * 6 m = 24 m.
Answer: 3.75 m
Drop 1 takes t where 5 = (1/2)(10)t^2 -> t = 1 s. Three drops at equal 0.5 s intervals, so drop 2 has fallen 0.5 s: distance = (1/2)(10)(0.5)^2 = 1.25 m, height = 5 - 1.25 = 3.75 m.
Answer: 4 s
Trains approach, so relative speed = 10+15 = 25 m/s. To fully cross, the relative displacement = sum of lengths = 50+50 = 100 m. Time = 100/25 = 4 s.
Answer: zero
From t = sqrt(x) + 3, sqrt(x) = t-3 so x = (t-3)^2 and v = dx/dt = 2(t-3). Velocity is zero at t = 3 s, at which x = (3-3)^2 = 0, so the displacement is zero.
Answer: 9 s
The correct option is 9 seconds because, using the equations of motion for uniformly accelerated motion, we can calculate the time taken to cover the distance of 135 m while the speed changes from 10 m/s to 20 m/s. The average speed during this interval is 15 m/s, and using the formula distance = average speed × time, we find that time equals 135 m / 15 m/s, which results in 9 seconds.
Answer: 80 m
For constant braking deceleration, d = v^2/(2a), so d is proportional to v^2. Doubling the speed (60 -> 120 km/h) multiplies the stopping distance by 4: 20 m x 4 = 80 m.
Answer: T = √(2h/g) + h/v
The correct option accounts for two phases: the time it takes for the stone to fall to the water surface, which is determined by the equation of motion under gravity, and the time for the sound to travel back up to the top of the well, which is based on the speed of sound. Thus, the total time is the sum of these two components.
Answer: 45 km/hr
The two slow cars are 5 km apart; the oncoming car closes this gap at relative speed (v+30) km/h. Meeting them 4 min (=1/15 h) apart gives 5/(v+30) = 1/15, so v+30 = 75, v = 45 km/h.
Answer: 16 m
x = 40 + 12t - t^3, so v = dx/dt = 12 - 3t^2 = 0 gives t = 2 s. Velocity stays positive for 0<t<2, so the particle moves monotonically; distance covered = x(2) - x(0) = (40 + 24 - 8) - 40 = 16 m.
Answer: 2400 m
Distance in nth second = u + a(n-1/2): u+4.5a=65 and u+8.5a=105 give a=10, u=20. In 20 s, s = 20*20 + (1/2)(10)(20^2) = 400 + 2000 = 2400 m.
Answer: 20: 7
Height for speed v is h=(u^2-v^2)/2g. With v = u/2, u/3, u/4 the heights are (3/4, 8/9, 15/16)(u^2/2g). AB = 8/9-3/4 = 5/36 and BC = 15/16-8/9 = 7/144 (in units of u^2/2g), giving AB:BC = (20/144):(7/144) = 20:7.
Answer: 160 minutes
In still water 8 km each way in 2 h gives boat speed 16/2 = 8 km/h. With current 4 km/h: upstream speed 4 km/h takes 8/4 = 2 h; downstream speed 12 km/h takes 8/12 = 2/3 h. Total = 2 + 2/3 h = 8/3 h = 160 minutes.
Answer: S2 = 4S1
Starting from rest with constant acceleration, displacement S = (1/2)a t^2, so S is proportional to t^2. Hence S2/S1 = (20/10)^2 = 4, giving S2 = 4 S1.
Answer: 3/5
The average velocity is calculated by dividing the total distance traveled by the total time taken. In this scenario, the particle covers a total distance of 6x and the time taken can be derived from the equations of motion for each segment, leading to an average velocity that is 3/5 of the maximum velocity achieved during the motion.
Answer: 10 s
To determine the time it takes for the parrot to pass the entire train, we first calculate the relative speed between the parrot and the train, which is the sum of their speeds since they are moving in opposite directions. The relative speed is 10 m/s (train) + 5 m/s (parrot) = 15 m/s. Then, we divide the length of the train (150 m) by the relative speed (15 m/s), resulting in 10 seconds.
Answer: S = 1/72 ft²
v^2 = 2fS so accel covers S; deceleration at f/2 covers v^2/f = 2S; cruise covers vt. Total = S + vt + 2S = 15S gives vt = 12S. Then t = 12S/sqrt(2fS) -> t^2 = 72S/f, so S = f t^2 / 72.
Answer: -2av³
The acceleration can be derived from the relationship between time and distance, where the second derivative of distance with respect to time gives acceleration. By differentiating the given equation twice, we find that the acceleration is proportional to the negative of the constant 'a' multiplied by the cube of the velocity, leading to the correct option.
Answer: t²
From v = dx/dt = a*sqrt(x): integral dx/sqrt(x) = integral a dt gives 2*sqrt(x) = a*t, so x = a^2*t^2/4. Displacement varies as t^2.
Answer: v0 + g/2 + f/3
The correct option is derived from integrating the velocity function to find the displacement. By evaluating the integral of the velocity equation from time 0 to 1, we find that the displacement is v0 + g/2 + f/3, which accounts for the contributions of both the linear and quadratic terms in time.
Answer: 2 s
dv/dt = -2.5*sqrt(v) -> integral v^(-1/2) dv = -2.5*integral dt -> 2*sqrt(v) = -2.5 t + C. At t=0, v=6.25 so C = 2*2.5 = 5. Setting v=0 gives 0 = -2.5 t + 5 -> t = 2 s.
Answer: 2gH = nu²(n − 2)
The correct option relates the height of the tower, the initial velocity, and the time taken for the particle to reach the ground and its highest point. By analyzing the motion of the particle, we can derive that the total time of flight is proportional to the square of the initial velocity and the height, leading to the equation 2gH = nu²(n − 2), which accounts for the time ratios involved.
Answer: g − a, g
The person inside the lift experiences a reduced effective gravitational acceleration due to the downward acceleration of the lift, resulting in the ball accelerating at g - a relative to them. However, an observer on the ground sees the ball accelerating at the full gravitational acceleration g, as they are not affected by the lift's motion.
Answer: 293 m
After free fall of 50 m: v^2 = 2g*50 ~ 980-1000. Then decelerating at 2 m/s^2 to 3 m/s: s = (v^2 - 9)/4 ~ 243 m. Total height ~ 50 + 243 = 293 m.
Answer: 2,000 J to 5,000 J
v = 100 m / 10 s = 10 m/s. With a sprinter mass of about 70 kg, KE = 0.5*70*10^2 = 3500 J, which lies in the 2,000 J to 5,000 J range.
Answer: 160 m
The stopping distance of a vehicle is proportional to the square of its speed. Therefore, if the speed doubles from 40 km/h to 80 km/h, the stopping distance increases by a factor of four, resulting in a new stopping distance of 160 m.
Answer: √(a1 a2) t
For equal race distance s: t_A=sqrt(2s/a1), t_B=sqrt(2s/a2), v_A=sqrt(2*a1*s), v_B=sqrt(2*a2*s). Then v=v_A-v_B and t=t_B-t_A give v/t = sqrt(a1*a2), so v = sqrt(a1*a2)*t.
Answer: (A), (B), (D)
Uniform acceleration from rest: a is constant positive (A); v = a t is a straight line from origin (B); x = (1/2)a t^2 is a parabola with increasing slope, concave up (D). So (A),(B),(D) are correct.
Answer: 1/√(γg) tan⁻¹(√(γ/g) V0)
Going up: m dv/dt = -mg - m*gamma*v^2. Then t = integral_0^{V0} dv/(g+gamma v^2) = (1/sqrt(gamma g)) * arctan(sqrt(gamma/g)*V0).
Answer: b²τ/2
v = b*sqrt(x) -> a = v(dv/dx) = b*sqrt(x) * b/(2*sqrt(x)) = b^2/2, a constant. Starting from rest at origin, v = a*t = (b^2/2)*tau = b^2*tau/2.
Answer: t = 3.4 √(h/g)
At height h the helicopter speed is v = sqrt(2gh) upward; the packet then has u=+sqrt(2gh), y0=h, a=-g. Solving h + sqrt(2gh) t - (1/2) g t^2 = 0 gives t = (2 + sqrt2) sqrt(h/g) ~ 3.4 sqrt(h/g).
Answer: αβ/(2(α + β)) t²
The correct option is derived from the kinematic equations for uniformly accelerated motion. The car accelerates and then decelerates, and the total distance can be expressed as the sum of the distances covered during both phases, leading to the formula αβ/(2(α + β)) t², which accounts for the average speed during the entire motion.
Answer: 2.50 m s⁻¹
The average speed of the ball can be calculated by considering the total distance traveled during its bounces and the total time taken. Since the ball bounces to a height of 4.05 m after the first drop and continues to rise to 81% of the previous height, the total distance is a geometric series, and the time can be derived from the heights using the formula for free fall. The resulting average speed comes out to be 2.50 m/s.
Answer: v₀ + g/2 + F/3
x(1) = integral_0^1 (v0 + g t + F t^2) dt = v0 + g/2 + F/3.
Answer: 200
The body undergoes a uniform angular acceleration, which allows us to use the formula for angular displacement during acceleration. By calculating the average angular speed and multiplying it by the time, we find that the total number of rotations made is 200.
Answer: (3v0²/2α)^(1/3)
With m v dv/dx = -alpha x^2, integrate: (1/2)v0^2 = (alpha/3m) x^3 (taking m as given), so the stopping distance x = (3 v0^2 / (2 alpha))^(1/3).
Answer: √((v² + u²)/2)
With uniform acceleration, v^2 = u^2 + 2aL over the full train. At the midpoint: v_mid^2 = u^2 + 2a(L/2) = u^2 + (v^2 - u^2)/2 = (u^2 + v^2)/2, so v_mid = sqrt((u^2 + v^2)/2).
Answer: 45 m
The first stone falls from the top and covers the distance of 45 m in a certain time, while the second stone, starting 25 m below the top, falls 20 m in the same time. Since both stones reach the bottom simultaneously, the total height of the building must be 45 m.
Answer: 2 mv³
From t=mx^2+nx, dt/dx=2mx+n=1/v. Then a=v dv/dx; dv/dx=-2m v^2, so a=-2m v^3. The retardation is 2m v^3.
Answer: 125 m
The dropped object has u = +10 m/s upward from 75 m: 75 + 10t - 5t^2 = 0 -> t^2 - 2t - 15 = 0 -> t = 5 s. The balloon keeps rising at 10 m/s, so its height = 75 + 10x5 = 125 m.
Answer: α/2 + β/3
To find the distance traveled between 1s and 2s, we need to integrate the velocity function v = αt + βt² over that interval. The correct option, α/2 + β/3, results from evaluating the definite integral of the velocity function from t=1 to t=2, which gives the total distance traveled.
Answer: a2/a1
The ratio of the times t1 and t2 is determined by the relationship between the acceleration and deceleration rates. Since the scooter accelerates at a1 and then decelerates at a2, the time spent accelerating (t1) is proportional to the acceleration rate, while the time spent decelerating (t2) is proportional to the deceleration rate, leading to the ratio t1/t2 = a2/a1.
Answer: 7.35 m
Time for the first drop to fall 9.8 m: t=sqrt(2h/g)=sqrt(2) s. Drops fall at equal interval T=t/2 (3rd starts as 1st lands). The 2nd drop has fallen for t/2 s: distance=0.5*9.8*(t/2)^2=2.45 m, so its height above the floor is 9.8-2.45=7.35 m.
Answer: (√3 − √2) / (√3 + √2)
With u=sqrt(2gh), the two times to reach h/3 satisfy g t^2-2u t+2h/3=0, giving t=[sqrt(2gh)+/-sqrt(4gh/3)]/g. The ratio of smaller to larger root simplifies to (sqrt3-sqrt2)/(sqrt3+sqrt2) (approx 0.101).
Answer: 1 drop / second
The correct option is 1 drop per second because the distance fallen by each droplet after 4 seconds, calculated using the formula for free fall (d = 0.5 * g * t²), matches the observed spacing of 34.3 m, indicating that a new droplet is released every second.
Answer: √2: √3
Ascent deceleration = (mg + f)/m = (50+10)/5 = 12 m/s^2; descent acceleration = (mg - f)/m = (50-10)/5 = 8 m/s^2. For the same height, t_ascent/t_descent = sqrt(a_descent/a_ascent) = sqrt(8/12) = sqrt(2):sqrt(3).