Chapter : 1. Motion
Examples on Body Falling Freely Under Gravity
Example : 10
A body starts moving with an initial velocity 50 m/s and acceleration 20 m/s2. How much distance it will cover in 4s ? Also, calculate its average speed during this time interval.
Solution. Given : u = 50 m/s, a = 20 m/s2, t = 4s, s = ?
S = ut +
at2 = 50 × 4 + × 20 × (4) 2
= 200 + 160 = 360 m
Average speed during this interval,
=
= 90 m/s
Example : 11
A body is moving with a speed of 20 m/s. When certain force is applied, an acceleration of 4 m/s2 is produced. After how much time its velocity will be 80 m/s ?
Solution. Given : u = 20 m/s, a = 4 m/s2, v = 80 m/s, t = ?
Using equation, v = u + at, we get
80 = 20 + 4 × t
or 4t = 80 – 20 = 60
or t = 15 s
Therefore, after 15 seconds, the velocity of the body will be 80 m/s.
Example : 12
A body starts from rest and moves with a constant acceleration. It travels a distance s1in first 10 s, and a distance s2 in next 10 s. Find the relation between s2 and s1.
Solution. Given : u = 0, t1 = 10 s
Distance travelled in first 10 seconds, is given by
s1 = ut + at2 = 0 +×a×(10)2 = 50a ...(1)
To calculate the distance travelled in next 10s, we first calculate distance travelled in 20 s and then subtract distance travelled in first 10 s.
s = ut +at2 = 0 + × a × (20) 2= 200a ...(2)
Distance travelled in 10th second interval,
s2 = s – s1 = 200a – 50a ...(3)
or s2 = 150a
Now,
or s2 = 3s1
A body starts moving with an initial velocity 50 m/s and acceleration 20 m/s2. How much distance it will cover in 4s ? Also, calculate its average speed during this time interval.
Solution. Given : u = 50 m/s, a = 20 m/s2, t = 4s, s = ?
S = ut +
at2 = 50 × 4 + × 20 × (4) 2= 200 + 160 = 360 m
Average speed during this interval,
=
= 90 m/s Example : 11
A body is moving with a speed of 20 m/s. When certain force is applied, an acceleration of 4 m/s2 is produced. After how much time its velocity will be 80 m/s ?
Solution. Given : u = 20 m/s, a = 4 m/s2, v = 80 m/s, t = ?
Using equation, v = u + at, we get
80 = 20 + 4 × t
or 4t = 80 – 20 = 60
or t = 15 s
Therefore, after 15 seconds, the velocity of the body will be 80 m/s.
Example : 12
A body starts from rest and moves with a constant acceleration. It travels a distance s1in first 10 s, and a distance s2 in next 10 s. Find the relation between s2 and s1.
Solution. Given : u = 0, t1 = 10 s
Distance travelled in first 10 seconds, is given by
s1 = ut +
at2 = 0 +×a×(10)2 = 50a ...(1)To calculate the distance travelled in next 10s, we first calculate distance travelled in 20 s and then subtract distance travelled in first 10 s.
s = ut +
at2 = 0 + × a × (20) 2= 200a ...(2)Distance travelled in 10th second interval,
s2 = s – s1 = 200a – 50a ...(3)
or s2 = 150a
Now,
or s2 = 3s1
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