a body moving with a uniform acceleration travels distances of 24m and 64m during the first two equal consecutive intervals of time each of duration 4s.determine the initial velocity and acceleration of moving body.
d = vi t + (1/2) a t^2
24 = vi (4) + (1/2) a (4)^2
88 = vi (8) + (1/2) a (8^2)
--------------------------------
48 = 8 vi + 16 a
88 = 8 vi + 32 a
----------------- subtract
-40 = -16 a
a = 2.5 m/s^2
24 = 4 vi + 20
vi = 1 m/s
check my arithmetc
a motor cyclist goes from station a to b at 40km/h and then returns from b to a 120km/h.calculate average speed,average velocity during the round trip
To solve this problem, we need to use the equations of motion. In this case, we can use the equation:
d = ut + (1/2)at^2
where:
- d is the distance traveled
- u is the initial velocity
- a is the acceleration
- t is the time
From the given information, we know that during the first interval of 4 seconds, the distance traveled is 24m. So, we can write the equation as:
24 = u(4) + (1/2)a(4)^2
Simplifying this equation gives us:
24 = 4u + 8a
Similarly, during the second interval of 4 seconds, the distance traveled is 64m. We can write the equation as:
64 = u(8) + (1/2)a(8)^2
Simplifying this equation gives us:
64 = 8u + 32a
Now we have a system of two equations with two unknowns (u and a). We can solve this system of equations to find the initial velocity and acceleration.
Let's solve the system of equations using the method of substitution:
From the first equation, we can rewrite it as:
4u = 24 - 8a
Now substitute this expression for 4u into the second equation:
64 = (24 - 8a)(8) + 32a
Simplifying this equation gives us:
64 = 192 - 64a + 32a
Combine like terms:
64 = 192 - 32a
Rearrange the equation:
32a = 192 - 64
Simplifying further:
32a = 128
Dividing both sides by 32 gives us:
a = 4
Now substitute the value of a into any of the earlier equations to find the value of u. Let's use the first equation:
24 = 4u + 8(4)
Simplifying this equation gives us:
24 = 4u + 32
Subtract 32 from both sides:
-8 = 4u
Dividing both sides by 4 gives us:
u = -2
Therefore, the initial velocity (u) is -2 m/s and the acceleration (a) is 4 m/s^2.
case 1:
from a to b velocity=40km/h=11.1m/s
let 11.1=a
case 2:
from b to a velocity=120kh/h=33.3m/s
let 33.3=b
avg.speed = b-a/2
= 33.3-11.1/2
= 22/2
= 11m/s
avg. speed= 11m/s=40km/h
avg.velocity=0
because avg. velocity is total displacement/time.
now the cyclist travels from a to b then from b to a, back to its original
position. so the avg.velocity is zero.