Describing Motion Around Us – Notes, MCQs, Quiz & Worksheet

This indicates that the object is moving with uniform velocity. A horizontal line means the velocity value is constant, and being above the time axis means it's moving in the positive direction.

A ball dropped in vacuum has its velocity changing (increasing) due to gravity, but the acceleration due to gravity is constant (uniform).

Yes, the direction of acceleration can be opposite to the direction of velocity. This occurs during deceleration or retardation, for example, when a car applies brakes and slows down, its acceleration is opposite to its direction of motion.

In uniform circular motion, the velocity is tangential to the circle and its direction continuously changes, even if speed is constant. The acceleration (centripetal acceleration) is always directed towards the center of the circle.

Using s = ((u+v)/2)*t. Here u=10 m/s, v=20 m/s, t=5 s. So, s = ((10+20)/2)*5 = (30/2)*5 = 15*5 = 75 m. Alternatively, find 'a' first: a=(20-10)/5 = 2 m/s². Then s = ut + 0.5at² = 10*5 + 0.5*2*5² = 50 + 25 = 75 m.

Displacement equals distance only when the object moves in a single straight line and does not reverse its direction. Any change in direction makes distance greater than displacement.

Yes, the athlete's motion is an example of accelerated motion. Although the speed is constant, the direction of velocity changes continuously at every point along the circular path, thus implying a change in velocity and hence acceleration.

A straight line indicates uniform rate of change. A negative slope means the velocity is decreasing, hence it represents uniform deceleration (retardation).

In uniform circular motion, the instantaneous velocity continuously changes its direction, making it a changing vector. However, the average velocity over a complete circle is zero, as the displacement for one full revolution is zero.

Assertion (A) is false because if speed (magnitude of velocity) varies, velocity itself must vary. Constant velocity implies both constant speed AND constant direction. Reason (R) is true.

The slope of a velocity-time graph (change in velocity / change in time) gives the acceleration of the object.

Using a = (v-u)/t. Here u=20 m/s, v=0 m/s, t=10 s. So, a = (0-20)/10 = -2 m/s². Deceleration is the magnitude of negative acceleration, hence 2 m/s².

At the highest point of its trajectory when thrown vertically upwards, the stone momentarily stops before falling back down, so its instantaneous velocity is zero.

Let the total distance be 2d. Time for first half: t1 = d/40. Time for second half: t2 = d/60. Total time = d/40 + d/60 = (3d+2d)/120 = 5d/120 = d/24. Average speed = Total distance/Total time = 2d / (d/24) = 48 km/h.

Acceleration is a vector quantity because it has both magnitude and direction. A change in either speed (magnitude) or direction of motion (or both) results in acceleration.

Objects thrown upwards fall back down due to the constant acceleration due to gravity (g) acting downwards. This acceleration continuously reduces the upward velocity until it becomes zero at the peak, and then increases the velocity in the downward direction.

First, convert 18 km/h to m/s: 18 * (1000/3600) = 5 m/s. Then, use a = (v-u)/t where u=0, v=5 m/s, t=5s. So, a = (5-0)/5 = 1 m/s².

The object is moving with uniform speed. A straight line in a distance-time graph indicates constant speed, and passing through the origin means it starts from rest at zero distance.

Using the first equation of motion, v = u + at. Here, u=0, v=19.6 m/s, t=2s. So, 19.6 = 0 + a*2, which gives a = 19.6/2 = 9.8 m/s².

The assertion is true, and the given reason perfectly explains this scenario. At the peak of vertical motion, the object momentarily stops (zero velocity) but is still under the influence of gravitational acceleration (g).

A decreasing slope on a distance-time graph signifies that the object's speed is decreasing, meaning it is undergoing deceleration or retardation.

Both statements are true. In free fall, the only significant force is gravity, which imparts a constant acceleration (g). This constant force is the reason for the uniform acceleration.

In uniform motion, an object covers equal distances in equal intervals of time, meaning its speed is constant. In non-uniform motion, an object covers unequal distances in equal intervals of time, implying its speed is changing (accelerated or decelerated).

The displacement is the area under the velocity-time graph. For a rectangle, Area = base * height = 10 s * 15 m/s = 150 m.

This is due to relative motion and our frame of reference. While we are in motion with the train, our brain perceives the train as stationary. Therefore, objects outside the train appear to move in the opposite direction relative to our (moving) frame of reference.

Stage 1 (acceleration): u=0, a=2 m/s², t=10 s. v = u+at = 0+2*10 = 20 m/s. s1 = ut+0.5at² = 0 + 0.5*2*10² = 100 m.
Stage 2 (constant velocity): v=20 m/s, t=10 s. s2 = v*t = 20*10 = 200 m.
Stage 3 (deceleration): u=20 m/s, v=0, a=-4 m/s². v²-u²=2as => 0²-20²=2*(-4)*s3 => -400 = -8*s3 => s3 = 50 m.
Total distance = s1+s2+s3 = 100 + 200 + 50 = 350 m.

1. A car moving at a constant speed in a straight line (uniform velocity). 2. An object at rest (zero velocity, zero acceleration).

Let 'n' be the total time of fall. Distance covered in 'n' seconds: Sn = 0.5 * g * n². Distance covered in (n-1) seconds: S(n-1) = 0.5 * g * (n-1)². Distance in last second: Sn - S(n-1) = 45.
0.5*10*n² - 0.5*10*(n-1)² = 45
5n² - 5(n² - 2n + 1) = 45
5n² - 5n² + 10n - 5 = 45
10n = 50 => n = 5 seconds.
Total height (total distance covered) = Sn = 0.5 * 10 * 5² = 5 * 25 = 125 m.

The object first moves in a positive direction with uniform speed, then momentarily comes to rest, and finally moves back in the opposite (negative) direction with the same uniform speed.

Average acceleration a = (v-u)/t. Here u=10 m/s, v=40 m/s, t=5 s. So, a = (40-10)/5 = 30/5 = 6 m/s².