Sections:

Motion in One Dimension, Page 18

Solved Problems using Kinematics Equations

  1. A car is initially moving with a velocity of 2 m/s. Then it accelerates uniformly at the rate of 0.4 m/s2 for 5 seconds. What is the distance covered by the car in 5 seconds?

    Solution:
    Given: initial velocity vi = 2 m/s, acceleration, a = 0.4 m/s2 and time, t = 5 seconds

    The distance, s, can be calculated by using the following kinematic equation.

s = vt + 1/2 at squared

Substituting the values of v, a and t we get,

s = 2 x 5 + 1/2 x 0.04 x 0.5 squared; s = 10 + 5; s = 15 m

  1. A truck starts from rest and attains a velocity of 4 m/s in 3 seconds. What is its acceleration?

    Solution:
    Given: vi = 0, vf = 4 m/s and t = 3 seconds.

    The acceleration can be found out by using the following kinematic equation.

v sub f = v sub i + at; 4 = 0 + a x 3; a = 4/3; a = 1.33 m/s squared

  1. A truck accelerates uniformly at the rate of 0.8 m/s2 for a distance of 50 m. If the truck started with an initial velocity of 4 m/s, what is its velocity at the end of 50 m?

    Solution:
    Given: a = 0.8 m/s2, s = 50 m and vi = 4 m/s

    The velocity can be found out by using the following kinematic equation.

V sub f squared = v sub i squared + 2as

Substituting the values of a, s and vi we get,

V sub f squared = v sub i squared + 2 as; V sub f squared = 4 squared + 2 x 0.8 x 50; V sub f squared = 16 + 80; V sub f squared = 96; V sub f = square root of 96; V sub f = 9.79 m/s