1-A particle moving along the x -axis has its velocity described by the function

Question

1-A particle moving along the x-axis has its velocity described by the function vx =2t2m/s, where t is in s. Its initial position is x0 = 2.6 m at t0 = 0 s

At 2.3 s , what is the particle's position?

At 2.3 s , what is the particle's velocity?

2-David is driving a steady 26.0 m/s when he passes Tina, who is sitting in her car at rest. Tina begins to accelerate at a steady 2.90 m/s2 at the instant when David passes.

How far does Tina drive before passing David?

What is her speed as she passes him?

Explanation / Answer

Part 1)
v(x) = 2*t^2
at t = 0
The velocity is 0 m/s

The velocity is defined as the change in position over time aka the velocity is the derivative of the position function.

int= integration
v(x) = dx/dt =2t^2
int dx = int 2t^2 dt
x(t) = 2t^3/3 + C

x(0) = 0 + C
c = 2.6

Hence, x(t) = 2t^3/3 + 2.6

At 2.3 sec
Position of particle = x(2.3) = 2*(2.3)^3/3 + 2.6 = 10.7113 m
Velocity of particle = v(2.3) = 2*(2.3)^2 = 10.58

Part 2)

Let us consider the time stamp at which David passes tina as 0.

Tina's velocity at time t can be define as = u + a*t = 0 + 2.9*t
= 2.9*t m/s
Distance travelled = ut+1/2*a*t^2 = 0 + 1/2*2.9*t^2 = 1/2*2.9*t^2

David's velocity at time t can be defined as = 26 m/s

Now
When they both meet again then the distance travelled will be same.
Hence,
26*t = 1/2*2.9*t^2
Equating this :
{{t == 0}, {t == 17.931}}

Hence they meet again after 17.931 sec

Distance travelled = 466.206m
Tina's Speed = 2.9*17.931 = 51.9999 m/s

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