Uniform Acceleration

Uniform acceleration is a type of motion where an object moves with a constant acceleration. This type of motion is characterized by a constant change in velocity over time. The following are some sample problems solved of uniform acceleration from SL Arora class 11:

Problem 1: A car accelerates from rest at a rate of 4 m/s^2. What is the velocity of the car after 5 seconds?

Solution:

• Initial Velocity (u) = 0 m/s


• Acceleration (a) = 4 m/s^2


• Time (t) = 5 s


• Final Velocity (v) = ?

Using the formula v = u + at, we can find the final velocity:

v = u + at
v = 0 + (4 m/s^2)(5 s)
v = 20 m/s

Therefore, the velocity of the car after 5 seconds is 20 m/s.

Problem 2: A ball is thrown vertically upward with a speed of 20 m/s. How high does it rise before it starts falling?

Solution:

• Initial Velocity (u) = 20 m/s


• Acceleration (a) = -9.8 m/s^2 (negative because the ball is moving against gravity)


• Final Velocity (v) = 0 m/s (at the highest point, velocity becomes 0)


• Time (t) = ?


• Maximum Height (h) = ?

Using the formula v = u + at, we can find the time taken by the ball to reach the highest point:

v = u + at
0 = 20 m/s + (-9.8 m/s^2)t
t = 2.04 s

Using the formula h = ut + 1/2 at^2, we can find the maximum height:

h = ut + 1/2 at^2
h = (20 m/s)(2.04 s) + 1/2 (-9.8 m/s^2)(2.04 s)^2
h = 20.4 m

Therefore, the ball rises to a height of 20.4 m before it starts falling.

Problem 3: A train starts from rest and accelerates at a rate of 3 m/s^2. How long does it take to reach a speed of 36 km/h?

Solution:

• Initial Velocity (u) = 0 m/s


• Acceleration (a) = 3 m/s^2


• Final Velocity (v) = 36 km/h = 10 m/s (converted from km/h to m/s)


• Time (t) = ?

Using the formula v = u + at, we can find the time taken by the train to reach a speed of 10 m/s:

v = u + at
10 m/s = 0 + (3 m/s^2)t
t = 3.33 s

Therefore, the train takes 3.33 seconds to reach a speed of

Class 11 physics motion and straight line example sl arora