[PDF] Download Problems Based on Equations of Motion for Class 9 Science
EQUATIONS OF MOTION
Table of Contents
When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration during motion and the distance covered by it in a certain time interval by a set of equations known as the equations of motion. There are three such equations. These are:
where u is the initial velocity of the object which moves with uniform acceleration a for time t, v is the final velocity, and s is the distance travelled by the object in time t.
Eq. (1) describes the velocity-time relation and Eq. (2) represents the position-time relation. Eq. (3), which represents the relation between the position and the velocity, can be obtained from Eqs. (1) and (2) by eliminating t. These three equations can be derived by graphical method.
Numerical Problems on Equations of Motion
Equations of Motion Practice Set 1
Apply $v=u+at$, $s=ut+\frac{1}{2}at^2$, and $v^2-u^2=2as$Answer: (a) $2\text{ m/s}^2$ (b) $10\text{ m/s}$ (c) $100\text{ m}$
Explanation: Initial velocity $u = 0$. Final velocity $v = 72\text{ km/h} = 72 \times \frac{5}{18} = 20\text{ m/s}$. Time $t = 10\text{ s}$.
(a) Acceleration:
(b) Average velocity (for uniform acceleration):
(c) Distance travelled:
Answer: 20 m
Explanation: Initial velocity $u = 0$, acceleration $a = 10\text{ m/s}^2$, time $t = 2\text{ s}$.
Using the second equation of motion:
Answer: $8.72 \times 10^{-5}\text{ m/s}$ (or $0.523\text{ cm/min}$)
Explanation: Radius $r = 5\text{ cm} = 0.05\text{ m}$. Time taken for one revolution by a minute hand is $T = 60\text{ minutes} = 3600\text{ s}$.
Answer: 4 m
Explanation: Initial velocity $u = 0$, $a = 8\text{ m/s}^2$, $t = 1\text{ s}$.
Answer: 40 m/s
Explanation: Final velocity $v = 0$ (brought to rest). Time $t = 20\text{ s}$. Retardation is negative acceleration, so $a = -2\text{ m/s}^2$.
Using the first equation of motion:
Answer: $1\text{ m/s}^2$
Explanation: Starting means $u = 0$. Final velocity $v = 36\text{ km/h} = 36 \times \frac{5}{18} = 10\text{ m/s}$. Time $t = 10\text{ s}$.
Answer: 200 m
Explanation: Initial velocity $u = 0$, $a = 4\text{ m/s}^2$, $t = 10\text{ s}$.
Answer: (i) $1\text{ m/s}^2$ (ii) $37.5\text{ m}$
Explanation: Convert speeds: $u = 18 \times \frac{5}{18} = 5\text{ m/s}$, $v = 36 \times \frac{5}{18} = 10\text{ m/s}$. Time $t = 5\text{ s}$.
(i) Acceleration:
(ii) Distance:
Answer: 10 s
Explanation: Initial velocity $u = 0.5\text{ m/s}$. Final velocity $v = 0$. Acceleration $a = -0.05\text{ m/s}^2$.
Answer: $\frac{5}{6}\text{ m/s}^2$ (or $\approx 0.833\text{ m/s}^2$)
Explanation: Change in velocity $v – u = 50 – 20 = 30\text{ km/h}$. Convert this to m/s: $30 \times \frac{5}{18} = \frac{150}{18} = \frac{25}{3}\text{ m/s}$. Time $t = 10\text{ s}$.
Equations of Motion Practice Set 2
Apply $v=u+at$, $s=ut+\frac{1}{2}at^2$, and $v^2-u^2=2as$Answer: $\frac{5}{9}\text{ m/s}^2$ (or $\approx 0.556\text{ m/s}^2$)
Explanation: Change in velocity $v – u = 58 – 56 = 2\text{ km/h}$. Convert to m/s: $2 \times \frac{5}{18} = \frac{5}{9}\text{ m/s}$. Time $t = 1\text{ s}$.
Answer: $60\text{ m/s}$ and $9000\text{ m}$ (or 9 km)
Explanation: $u = 0$, $a = 0.2\text{ m/s}^2$, $t = 5 \times 60 = 300\text{ s}$.
Speed acquired:
Distance travelled:
Answer: $a = -1.875\text{ m/s}^2$, Distance = 60 m
Explanation: $u = 54 \times \frac{5}{18} = 15\text{ m/s}$. Final velocity $v = 0$. Time $t = 8\text{ s}$.
Acceleration:
Distance:
Answer: (i) $\frac{1}{15}\text{ m/s}^2$ (ii) $3000\text{ m}$
Explanation: $u = 0$, $v = 72 \times \frac{5}{18} = 20\text{ m/s}$, $t = 5 \times 60 = 300\text{ s}$.
(i) Acceleration:
(ii) Distance:
Answer: $0.157\text{ cm/s}$
Explanation: Radius $r = 1.5\text{ cm}$. A second’s hand completes one revolution in $T = 60\text{ s}$.
Answer: $1.1\text{ m/s}$
Explanation: Diameter $D = 105\text{ m}$. Time $t = 5 \times 60 = 300\text{ s}$.
Answer: (i) $1.31\text{ m/s}$ (ii) $0\text{ m/s}$
Explanation: Radius $r = 50\text{ m}$. Time $t = 4 \times 60 = 240\text{ s}$.
(i) Distance for one revolution $= 2\pi r = 2 \times 3.14 \times 50 = 314\text{ m}$.
(ii) After one full revolution, displacement is zero. Hence, average velocity is $0\text{ m/s}$.
Answer: 320 m
Explanation: The motion happens in two phases.
Phase 1 (First 8s): $u=0$, $a=5\text{ m/s}^2$, $t=8\text{ s}$.
Velocity acquired after 8 seconds: $v = u + at = 0 + 5(8) = 40\text{ m/s}$.
Phase 2 (Remaining 4s): Travels with constant velocity $40\text{ m/s}$.
Answer: $v = 7\text{ m/s}$, Distance = 60 m
Explanation: $u = 5\text{ m/s}$, $a = 0.2\text{ m/s}^2$, $t = 10\text{ s}$.
Final Speed:
Distance:
Answer: 12 m
Explanation: Acceleration $a = -6\text{ m/s}^2$, Time $t = 2\text{ s}$, Final velocity $v = 0$. First, find initial velocity $u$:
Now find distance:
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Importance of Solving Numericals for Motion Class 9
Solving numerical problems in Class 9 plays a crucial role in the learning process and is essential for several reasons:
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- Exam preparation: Many examinations, including Class 9 assessments, include numerical questions. Practicing numerical problems prepares students for these exams and helps them perform well.
- Building logical thinking: Numerical problems often require a logical approach to arrive at the correct solution. By practicing such problems, students enhance their logical reasoning skills.
- Retention of concepts: Actively engaging with numerical problems reinforces the concepts learned in class, improving retention and recall during exams and beyond.
- Connecting theory with practice: Numerical problem-solving bridges the gap between theoretical knowledge and its practical applications. This connection helps students grasp the real significance of the concepts they are studying.
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