A body moving with an initial velocity of 10m/s has a uniform acceleration of 2m/s.Find distance travelled in the 8th second of its motion?
To find the distance traveled in the 8th second of motion, we need to use the equations of motion. Let's break down the information given:
Initial velocity (u) = 10 m/s
Acceleration (a) = 2 m/s²
Time (t) = 8 s
We can use the equation of motion that relates displacement (S), initial velocity (u), acceleration (a), and time (t):
S = ut + (1/2)at²
First, let's find the final velocity (v) at the end of the 8th second using the equation:
v = u + at
Substituting the given values:
v = 10 m/s + 2 m/s² * 8 s
v = 10 m/s + 16 m/s
v = 26 m/s
Now, we can calculate the displacement (S) during the 8th second:
S = ut + (1/2)at²
S = 10 m/s * 8 s + (1/2) * 2 m/s² * (8 s)²
S = 80 m + (1/2) * 2 m/s² * 64 s²
S = 80 m + 2 m/s² * 32 s²
S = 80 m + 64 m
S = 144 m
Therefore, the distance traveled in the 8th second of motion is 144 meters.
s(t) = 10t + t^2
s(7) = 70+49 = 119
s(8) = 80 + 64 = 144
distance in 8th second = 144-119 = 25m
or, s = ∫[7,8] 10+2t dt