A body with uniform acceleration covers 76 m in the 5th sec and 116 m in the 10th sec.Find the distance traveled in the 15th sec.what is the distance traveled in next 5 seconds.

As the body is travelling with uniform acceleration so the distance traveled by it would in 15th sec would be 156 m but how to calculate the distance in next 5 seconds

To find the distance traveled in the next 5 seconds, we need to determine the acceleration of the body.

Given that the body traveled 76 m in the 5th second and 116 m in the 10th second, we can calculate the acceleration using the formula:

Acceleration (a) = (final velocity - initial velocity) / time

For the 5th second:
Distance = (initial velocity × time) + (0.5 × acceleration × time^2)
76 = (0 × 5) + (0.5 × a × 5^2)
76 = 12.5a

For the 10th second:
Distance = (initial velocity × time) + (0.5 × acceleration × time^2)
116 = (0 × 10) + (0.5 × a × 10^2)
116 = 50a

Solving these two equations simultaneously, we can find the acceleration:

12.5a = 76, a = 6.08 m/s^2
50a = 116, a = 2.32 m/s^2

Since these two values are not equal, there seems to be an error. Please double-check the given information.

To calculate the distance traveled in the next 5 seconds, we need to determine the acceleration of the body first.

Given that the body covers 76 m in the 5th second and 116 m in the 10th second, we can find the change in velocity (Δv) for this interval.

Δv = Final velocity - Initial velocity

Using the formula, we have:
Δv = (116 m) - (76 m)
Δv = 40 m/s

Now, we can calculate the acceleration (a) by dividing the change in velocity by the time interval.

a = Δv / Δt

For the given time interval of 5 seconds, we have:
a = 40 m/s / 5 s
a = 8 m/s²

Now that we know the acceleration, we can find the distance traveled in the next 5 seconds by using the formula for distance covered with uniform acceleration:

distance (d) = initial velocity (v) * time (t) + 0.5 * acceleration (a) * time²

Given that the initial velocity is the velocity at the end of the 10th second, we can calculate v using the formula:

v = u + a * t

Here, u is the velocity at the end of the 5th second, which is the initial velocity before the 6th second.

By substituting the values into the formula, we have:

v = (u) + (a * Δt)

v = (116 m) + (8 m/s² * 5 s)
v = (116 m) + (40 m)
v = 156 m/s

Now we can substitute the values into the distance formula:

distance (d) = (156 m/s) * (5 s) + 0.5 * (8 m/s²) * (5 s)²

Calculating this, we get:

d = 780 m + 0.5 * 8 m/s² * 25 s²
d = 780 m + 100 m
d = 880 m

Therefore, the distance traveled in the next 5 seconds is 880 meters.

Ghgh

I would do it this way

find acceleration, and initial velocity.

distance covered in the fifth second is distance in the first five seconds minus distance in the first four...
d(5th)=vi(5-4)+a/2 (25-16)=vi+4.5a
76=vi+4.5a
now, the 10th second
116=vi(10-9)+a/2(100-81) or
116=vi+9.5a
so you have two equations, two unknowsn, solve for vi, and a.

Now, for the distance in the 16thru 21 seconds
d=vi(20-15)+a/2 (20^2-15^2)