An object has velocity u m/s at time 0 and constant acceleration a.
a. Use the fundamental theorem of calculus to find the velocity after t seconds.
v(t) =
m
/s
b. Use the fundamental theorem a second time to find the displacement of the object between time 0 and time t.
s(t) =
meters

a. The expression of velocity is,

b. The expression of displacement is,

Acceleration is defined as the rate of change of velocity with respect to time.

Integrate both side.

Given that, at time t = 0, v = u

Substitute ,  

So that expression of velocity is,

Velocity is defined as the rate of change of displacement with respect to time.

Integrate on both side.

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Part A) v(t) = u + a t   m/sec

Part B) s(t) = s₀ + ut + 0.5 at²   meters

Step-by-step explanation:

Given: An object has velocity u m/s at time 0 and constant acceleration a.

Part A) find the velocity after t seconds.

We should know that:

By integrating both sides with respect to the time t

∴∫ dv = ∫ a dt

∴ v(t) = a t + constant

the object has velocity u m/s at time 0 ⇒ constant = u

∴ v(t) = u + a t  m/sec

Part B) find the displacement of the object between time 0 and time t.

We should know that:

By integrating both sides with respect to the time t

∴∫ ds = ∫ v dt

∵ v(t) = u + a t

∴ ∫ ds = ∫ (u + a t) dt

∴ s(t) = ut + 0.5 at² + constant

Let at t = 0 displacement = s₀ ⇒ ∴ Constant = s₀

∴ s(t) = s₀ + ut + 0.5 at²  meters

10.416 meters per second

step-by-step explanation:

distance=1 00 meters

time= 9.6 seconds in decimal form or 9 3/5 as a mixed number

average speed= 100/9.6

=10.416 meters per second or 10 52/125 as a mixed number.

0.32

step-by-step explanation:

2.24/7