Find the distance, in meters, a particle travels in its first 10 seconds of travel, if it moves according to the velocity equation v(t)= 49 − 9.8t (in meters/sec)

a) -49
b)0
c) 245
d) 441

answer c= 245???

To find the distance a particle travels in the first 10 seconds, we need to find the total displacement during that time interval.

We can do this by integrating the velocity equation with respect to time from t = 0 to t = 10:

∫[0 to 10] (49 - 9.8t) dt

This can be simplified as follows:

= ∫[0 to 10] 49 dt - ∫[0 to 10] 9.8t dt
= 49t - (4.9t^2) [0 to 10]
= 49(10) - 4.9(10^2) - 49(0) + 4.9(0^2)
= 490 - 490
= 0

Therefore, the particle does not travel any distance in the first 10 seconds of travel. So, the correct answer is option b) 0.

To find the distance that a particle travels in its first 10 seconds, we need to calculate the integral of the velocity equation over the interval [0, 10].

The integral of the velocity function v(t) with respect to time will give us the displacement function s(t). Since the particle's initial position is not given, we need to calculate the definite integral from t = 0 to t = 10.

The integral of v(t) = 49 - 9.8t with respect to t is:
S(t) = ∫(49 - 9.8t) dt

To find the displacement function, integrate the first term and second term separately:
S(t) = ∫49 dt - ∫9.8t dt
S(t) = 49t - (9.8/2)t² + C (where C is the constant of integration)

Now, evaluate the displacement function S(t) at t=10 and t=0 to find the distance traveled by the particle in the first 10 seconds:

S(10) - S(0) = 49(10) - (9.8/2)(10²) - (0) = 490 - 490 = 0.

Therefore, the correct answer is b) 0. The particle does not travel any distance in its first 10 seconds of travel.

B

integrate from 0..10 49-9.8t dt = 49t-4.9t^2