Test: Kinematic Equations for Uniformly Accelerated Motion - ACT MCQ

In uniformly accelerated motion, at least three variables are required to completely define the system This can also be found out by using the equations of motion. For finding out the value of any variable, we need at least three known quantities.

Apply the first equation of motion in both the parts. For first part, v₁ = u + a₁t₁ = 0 + 5×10 = 50 m/s. For second part, v₂ = u₂ (= v₁) + a₂t₂ = 50 + 15×20 = 350 m/s.

The formula for conversion is, km/h = (18/5) x m/s. Therefore, 15 m/s = 15 x (18/5) = 54 km/h.

Average acceleration is the total change in velocity by the total change in time. Here, total change in velocity = 50 - 10 = 40m/s, and total change in time = 30 + 10 = 40s. Therefore, average acceleration = 1 m/s².

Here we will use the third equation of motion. The third equation of motion is v² = u² + 2as. At the maximum height, the velocity of the ball will be zero, therefore, v = 0, u = 20, and a = -g = -10. On solving, we will get s = 20m.

There are primarily 3 equations of motion. These are v = u + at, s = ut + (1/2)at² and v² = u² + 2as.

The three equations of motion are valid for uniformly accelerated motion. The equations do not work in situations where the acceleration is non-uniform. In that case it is better to work with the differential forms of velocity and acceleration.

Under the application of no external force like gravity, the acceleration is zero. Hence, both the objects will move with the same initial velocity.

Use the second equation of motion to find the acceleration. s = ut + (1/2)at², u = 0, t =15, s=15, therefore, a = 2/15 m/min². Now use the first equation to find the final velocity. v = u + at, u = 0, t = 15 min, a = 2/15, therefore, v = 2 m/min.

We can use the first equation of motion here. The first equation says, v = u + at, here, u = 1 m/h, t = 10 Hrs, a = 0.1 m/h². On putting all the values in the equation, we will get v = 2 m/h.