1. A body is moving with uniform acceleration. And describes 75m in the 6th second and 115 m in the 11th second. Show that it moves 155m during the 16th second.
2. A body is dropped from the roof of a building 10m high. Calculate the time of fall, and the speed with which it hits the ground. Take g=10m/s^2
Answers
Answered by
40
Formula used: Sₓ=u+a(n-1/2)
To find the acceleration:
According to the given data at two positions,
the equations we get using the formula and on solving them, we will get
-75 = -u - (11/2)a → 1 multiplying with (-) on both sides
115 = u + (21/2)a → 2
______________
40 = 0 + 5a
______________
Then, acceleration a = 8 m/s² in 1 (eliminating (-))
We get 75 = u + (11/2)8 ⇒ u = 75-44 = 31 m/s
So, initial velocity, u = 31 m/s
So, at 16th second:
S₁₆ = 31 + (16-1/2)8
= 31 + 31x4
= 31+124
= 155 m
Therefore, at 16th second, the body will be at 155 m. Hence, it is proved. ========================
To find the acceleration:
According to the given data at two positions,
the equations we get using the formula and on solving them, we will get
-75 = -u - (11/2)a → 1 multiplying with (-) on both sides
115 = u + (21/2)a → 2
______________
40 = 0 + 5a
______________
Then, acceleration a = 8 m/s² in 1 (eliminating (-))
We get 75 = u + (11/2)8 ⇒ u = 75-44 = 31 m/s
So, initial velocity, u = 31 m/s
So, at 16th second:
S₁₆ = 31 + (16-1/2)8
= 31 + 31x4
= 31+124
= 155 m
Therefore, at 16th second, the body will be at 155 m. Hence, it is proved. ========================
Answered by
1
Answer:
If initial velocity be u and the acceleration be a then the distance in first 5s will be S
5 =u(5)+ 2a(5) 2
=40m........(1)and the distance in first 10s will be S
10 =u(10)+2a(10) 2
so the distance between time t=5s to t=10 will be S=S
10 −S
5 =65m
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