Here are 5 problems based on equations of motion along with their detailed solutions. These are helpful for class 9–10 level students.
Equation of Motion Used:
- V = u + a.t
- S = u.t + ½ a.t2
- V2 = U2 + 2.a.s { where V = Final velocity , u = initial Velocity , t = time period and S = distance travelled }
Problem 1:
A car starts from rest and accelerates uniformly at . What is its velocity after 5 seconds?
Solution:
Given: u= 0 , time T = 5 sec
Using:
v = 0 + (4)(5) = 20 m/s
Answer:
Problem 2:
A train moving with an initial velocity of is brought to rest by applying brakes. If the retardation is , find the distance it travels before stopping.
Solution:
Given:
,
,
Using:
0 = 30^2 + 2(-3)s \Rightarrow 0 = 900 – 6s \Rightarrow s = \frac{900}{6} = 150 \, m
Answer:
Problem 3:
A body starts from rest and travels 80 m in 4 seconds under uniform acceleration. Find the acceleration.
Solution:
Given:
,
,
Using:
80 = 0 + \frac{1}{2}a(4^2) \Rightarrow 80 = 8a \Rightarrow a = \frac{80}{8} = 10 \, m/s^2
Answer:
Problem 4:
A bullet moving at hits a wall and comes to rest in . Find the retardation.
Solution:
Given:
,
,
Using:
0 = 200 + a(0.01) \Rightarrow a = \frac{-200}{0.01} = -20,000 \, m/s^2
Answer: (retardation)
Problem 5:
A car covers 100 m in 5 seconds with a uniform acceleration starting from . Find the acceleration.
Solution:
Given:
,
,
Using:
100 = 10(5) + \frac{1}{2}a(25) \Rightarrow 100 = 50 + 12.5a \Rightarrow 12.5a = 50 \Rightarrow a = 4 \, m/s^2
Answer:
–
Great! Here are 5 more problems on equations of motion with clear step-by-step solutions:
Problem 6:
A car accelerates from to in 4 seconds. Find the acceleration and the distance covered during this time.
Solution:
Given: t = 4 sec
Find out ; a = ? , d = ?
(i) Acceleration:
Using: formula V = U + a.t
25 = 15 + a(4)
a = {10}/{4} = 2.5 m/s^2
(ii) Distance covered:
Using:
s = 15(4) + {1}/{2}(2.5)(4^2) = 60 + {1}{2}(2.5)(16) = 60 + 20 = 80 m,
Answer: Acceleration = , Distance =
Problem 7:
An object moving with uniform acceleration travels 54 m in 3 seconds and 98 m in 5 seconds. Find the initial velocity and acceleration.
Given – distance = 54 m, Time t = 3 sec ,
Find out : initial velcoty , U = 0 , acceleration a = ?
Solution:
Let initial velocity and acceleration.
Using:
For 3 seconds:
→ Equation (1)
For 5 seconds:
→ Equation (2)
Now,
(1) →
(2) →
Multiply (1) by 5 and (2) by 3:
Subtract:
Substitute in (1):
Answer:
Problem 8:
A stone is dropped from a height of 80 m. How long will it take to reach the ground? (Take )
Solution:
Given:
,
,
Using:
80 = 0 + {1}{2}(9.8)t^2 , 80 = 4.9t^2 t^2 = \frac{80}{4.9} \approx 16.33 \Rightarrow t \approx 4.04 \, s
Answer:
Problem 9:
A vehicle moving at applies brakes and stops after 10 seconds. Find the distance it traveled after applying brakes.
Solution:
Given:
,
,
Using:
s = \frac{20 + 0}{2} \times 10 = 10 \times 10 = 100 \, m
Answer:
Problem 10:
A ball thrown vertically upwards reaches a maximum height of 45 m. Find the initial velocity. (Take )
Solution:
At max height, ,
,
Using:
0 = u^2 + 2(-9.8)(45) \Rightarrow u^2 = 882 \Rightarrow u = \sqrt{882} \approx 29.7 \, m/s
Answer:
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