JEE MainPhysicsMotion in A Straight LineMCQ+4 / −1
Train A and train B are running on parallel tracks in the opposite directions with speeds of 36 km/hour and 72 km/hour, respectively. A person is walking in train A in the direction opposite to its motion with a speed of 1.8 km/ hour. Speed (in ms–1) of this person as observed from train B will be close to : (take the distance between the tracks as negligible)
- A30.5 ms–1
- B29.5 ms–1
- C31.5 ms–1
- D28.5 ms–1
More from Motion in A Straight Line
- A Tennis ball is released from a height h and after freely falling on a wooden floor it rebounds and reaches height . The velocity versus height of the ball during its motion may be represented graphically by : (graph are…2020 · MCQ
- The speed verses time graph for a particle is shown in the figure. The distance travelled (in m) by the particle during the time interval t = 0 to t = 5 s will be . Includes diagram2020 · Numerical
- A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet is dropped from the helicopter when it is at a height h. The time taken by the packet to reach the ground is close to : [g is the…2020 · MCQ
- The velocity (v) and time (t) graph of a body in a straight line motion is shown in the figure. The point S is at 4.333 seconds. The total distance covered by the body in 6 s is : Includes diagram2020 · MCQ
- The distance x covered by a particle in one dimensional motion varies with time t as x2 = at2 + 2bt + c. If the acceleration of the particle depends on x as x–n, where n is an integer, the value of n is 2020 · Numerical
- A particle starts from origin O from rest and moves with a uniform acceleration along the positive x-axis. Identify all figures that correctly represent the motion qualitatively. (a = acceleration, v = velocity, x = displacement, t = time) Includes diagram2019 · MCQ
- The position vector of a particle changes with time according to the relation What is the magnitude of the acceleration at t = 1 ?2019 · MCQ
- The position of a particle as a function of time t, is given by x(t) = at + bt2 – ct3 where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be :2019 · MCQ