There are three basic equations of motion for bodies moving with uniform acceleration. These equations relate initial velocity, final velocity, acceleration, time, and distance covered by a moving body. To simplify the derivation of these equations, we assume that the motion is along a straight line. Hence, we consider only the magnitude of displacements, velocities, and acceleration.
Consider a body moving with initial velocity vi in a straight line with uniform acceleration a. Its
Velocity becomes v, after time t. The motion of the body is described by the speed-time graph as shown in figure 2.26 by line AB. The slope of line AB is acceleration a. The total distance covered by the body is shown by the shaded area under the line AB. Equations of motion can be obtained easily from this graph.
Speed-time graph for the motion of a body is shown in figure 2.26. The slope of line AB gives the acceleration a of a body.
Slope of line AB = a = BC / AC = BD - CD / OD as BD, CD = v, and OD = 1. Hence,
a = (V - Vi) / t
(2.4) ........... (2.5)
In the speed-time graph shown in figure 2.26, the total distance S traveled by the body is equal to the total area OABD under the graph. That is,
Total distance S = area of (rectangle OACD + triangle ABC)
Area of rectangle OACD = OA x OD
Area of triangle ABC = 1/2 (AC x BC) = 1/2 x a x t
Since Total area = area of rectangle OACD + area of triangle ABC
A train slows down from 80 km/h with a uniform retardation of 2 m/s². How long will it take to attain a speed of 20 km/h?
SOLUTION:
Using the equation V = Vinitial + at, we get:
5.6 m/s = 22.2 m/s + (-2 m/s²) × t
Solving for t, we get t = 8.3 s.
Thus, the train will take 8.3 s to attain the required speed.
A bicycle accelerates at 1 m/s² from an initial velocity of 4 m/s for 10 s. Find the distance moved by it during this interval of time.
SOLUTION:
Applying the equation for distance:
S = v0 × t + 0.5 × a × t²
S = 4 m/s × 10 s + 0.5 × 1 m/s² × (10 s)²
S = 40 m + 0.5 × 1 × 100
S = 40 m + 50 m = 90 m
Thus, the distance moved by the bicycle is 90 meters.
1 m/s = 3.6 km/h1 km/h = 1000 m / 3600 s = 0.27778 m/sA bicycle accelerates at 1 m/s² from an initial velocity of 4 m/s for 10 s. Find the distance moved by it during this interval of time.
SOLUTION
S = v0 × t + ½ × a × t²
= 4 m/s × 10 s + ½ × 1 m/s² × (10 s)²
= 40 m + 50 m
= 90 m
Thus, the bicycle will move 90 metres in 10 seconds.
A car travels with a velocity of 5 m/s. It then accelerates uniformly and travels a distance of 50 m. If the velocity reached is 15 m/s, find the acceleration and the time to travel this distance.
SOLUTION
Putting values in the third equation of motion, we get:
vfinal² = vinitial² + 2aS
(15 m/s)² = (5 m/s)² + 2a × 50 m
225 = 25 + 100a
200 = 100a
a = 2 m/s²
Using the first equation of motion to find t:
vfinal = vinitial + at
15 m/s = 5 m/s + 2 m/s² × t
15 m/s - 5 m/s = 2 m/s² × t
10 m/s = 2 m/s² × t
t = 5 s
Thus, the acceleration of the car is 2 m/s² and it takes 5 seconds to travel 50 m distance.
In the speed-time graph shown in figure 2.26, the total distance S travelled by the body is given by the total area OABD under the graph.
Total area OABD = (OA + BD) × OD / 2
S = (OA + BD) × OD / 2
Multiplying both sides by 2S:
2S = (OA + BD) × OD
Putting the values in the above equation 2.7, we get:
2S = (v + v0) × (v - v0) / 2a
2S = (v² - v0²) / 2a
A car travelling at 10 m/s accelerates uniformly at 2 m/s². Calculate its velocity after 5 s.
SOLUTION:
Using the equation for velocity with uniform acceleration:
V = vinitial + at
where:
Substituting the values, we get:
V = 10 m/s + 2 m/s² × 5 s
V = 10 m/s + 10 m/s
V = 20 m/s