A body moving with a uniform acceleration covers 0.65m in the fifth second and 1.05 m in the ninth second. What distance will it move in 15seconds?
Answers
Answer:
14.25 m
Step-by-step explanation:
A body moving with a uniform acceleration covers 0.65m in the fifth second and 1.05 m in the ninth second. What distance will it move in 15seconds?
Let say initial speed = u m/s
acceleration = a m/s²
S = ut + (1/2)at² S = Distance
Distance covered in 5th Sec = Distance covered in 5 sec - Distance covered in 4 sec
=>Distance covered in 9th Sec = 5u + (1/2)a5² - (4u + (1/2)a4²)
= u + 9a/2
u + 9a/2 = 0.65
=> 2u + 9a = 1.3 eq 1
Distance covered in 9th Sec = Distance covered in 9 sec - Distance covered in 8 sec
=>Distance covered in 9th Sec = 9u + (1/2)a9² - (8u + (1/2)a8²)
= u + 17a/2
u + 17a/2 = 1.05
=> 2u + 17a = 2.1 eq 2
Eq 2 - eq 1
=> 8a = 0.8
=> a = 0.1
acceleration = 0.1 m/s²
2u + 17*0.1 = 2.1
=> 2u = 2.1 - 1.7
=> u = 0.2
Initial Velocity = 0.2 m/s
Distnace in 15 seconds = 15u + (1/2)a(15)²
= 15 * 0.2 + (1/2)(0.1)225
= 3 + 11.25
= 14.25 m
Distance moved in 15 seconds = 14.25 m
And in 15th second = (15u + (1/2)a(15)²) - (14u + (1/2)a(14)²)
= u + 29a/2 = 0.2 + 14.5*.1 = 1.65 m
Answer:
h8
Step-by-step explanation: