Physics PDF - Acceleration Calculations

attain a velocity of 6 m s–1 in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m s-1 in the next 5 s. Calculate the acceleration of the bicycle in both the cases. To answer such a question, we have to introduce another physical quantity called acceleration, which...

attain a velocity of 6 m s–1 in 30 s. Then he applies brakes such that the velocity of the bicycle comes down to 4 m s-1 in the next 5 s. Calculate the acceleration of the bicycle in both the cases. To answer such a question, we have to introduce another physical quantity called acceleration, which is a measure of the change in the velocity of an object per unit time. That is, Solution: change in velocity acceleration = time taken In the first case: initial velocity, u = 0 ; final velocity, v = 6 m s–1 ; time, t = 30 s. From Eq. (8.3), we have If the velocity of an object changes from an initial value u to the final value v in time t, the acceleration a is, v–u a= t v–u (7.3) a= t This kind of motion is known as accelerated motion. The acceleration is taken to be positive if it is in the direction of velocity and negative when it is opposite to the direction of velocity. The SI unit of acceleration is m s–2. If an object travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, then the acceleration of the object is said to be uniform. The motion of a freely falling body is an example of uniformly accelerated motion. On the other hand, an object can travel with non-uniform acceleration if its velocity changes at a non-uniform rate. For example, if a car travelling along a straight road increases its speed by unequal amounts in equal intervals of time, then the car is said to be moving with non-uniform acceleration. Substituting the given values of u,v and t in the above equation, we get (6 m s a= –1 – 0m s –1 ) 30 s = 0.2 m s–2 In the second case: initial velocity, u = 6 m s–1; final velocity, v = 4 m s–1; time, t = 5 s. (4 m s Then, a = –1 – 6m s –1 ) 5s = –0.4 m s–2. The acceleration of the bicycle in the first case is 0.2 m s–2 and in the second case, it is –0.4 m s–2. Activity ______________ 7.8 In your everyday life you come across a range of motions in which (a) acceleration is in the direction of motion, (b) acceleration is against the direction of motion, (c) acceleration is uniform, (d) acceleration is non-uniform. Can you identify one example each for the above type of motion? Q Example 7.4 Starting from a stationary position, Rahul paddles his bicycle to uestions 1. When will you say a body is in (i) uniform acceleration? (ii) nonuniform acceleration? 2. A bus decreases its speed from 80 km h–1 to 60 km h–1 in 5 s. Find the acceleration of the bus. 3. A train starting from a railway station and moving with uniform acceleration attains a speed 40 km h–1 in 10 minutes. Find its acceleration. 77 MOTION Rationalised 2023-24