Question

The acceleration function (in m/s

2
) and the initial velocity v(0 ) (in m/s ) are given for a particle moving along a line. a(t)=2t+4,v(0)=−5,0≤t≤3 (a) Find the velocity (in m/s ) at time t. v(t)=m/s (b) Find the distance traveled (in m ) during the given time interval. ×m

Answer 1

(a) To find the velocity at time t, we need to integrate the acceleration function with respect to time.

∫a(t)dt = ∫(2t + 4)dt = t^2 + 4t + C

We can find the constant of integration by using the initial velocity, v(0) = -5.

v(0) = -5 = C

So the velocity function is:

v(t) = t^2 + 4t - 5

Plugging in t=3, we get:

v(3) = 3^2 + 4(3) - 5 = 16 m/s

Therefore, the velocity at time t is 16 m/s.

(b) To find the distance traveled during the given time interval, we need to integrate the velocity function with respect to time.

∫v(t)dt = ∫(t^2 + 4t - 5)dt = (1/3)t^3 + 2t^2 - 5t + C

We can find the constant of integration by using the initial position, x(0) = 0.

x(0) = 0 = C

So the position function is:

x(t) = (1/3)t^3 + 2t^2 - 5t

Plugging in t=3 and t=0, we get:

x(3) = (1/3)(3)^3 + 2(3)^2 - 5(3) = 9 m

x(0) = 0

Therefore, the distance traveled during the given time interval is 9 m.