Question

An automobile travels along the circular path of radius r = 35 m such that its speed is increased by a = (1.5e^t) m/s^2, where t is in seconds. Determine the magnitude of its velocity after it has traveled s = 20 m starting from rest. Neglect the size of the automobile.

Answer 1

15.678 m/s

Step-by-step explanation:

To solve this problem, we can use the kinematic equation:

⇒
`V^2=V_0^2+2a\triangle s`

where
`V` is the final velocity,
`V_0` is the initial velocity (which is zero in this case),
`A` is the acceleration, and
`\triangle s` is the displacement.

We know that the displacement of the car is Δs = 20 m, and the radius of the circular path is r = 35 m. We can use the relationship between displacement, radius, and angle (Δs = rΔθ) to find the angle through which the car has traveled:

⇒ t = sqrt((2 * 20 m) / (1.5 e^(0.867 s))) = 1.202 s

Using this value for the time in the kinematic equation:

• v^2 = v0^2 + 2aΔs
• v^2 = 0 + 2(1.5e^1.202)(20)
• v = 15.678 m/s

Therefore, the magnitude of the velocity of the car after it has traveled 20 m starting from rest along a circular path of radius 35 m is 15.678 m/s.

Answer 2

`15.678\; {\rm m\cdot s^(-1)}`.

Step-by-step explanation:

In this question, the rate of change in velocity is given. The speed of the vehicle after travelling the given distance can be found through the following steps:

• Find an expression for speed by Integrating the rate of change of speed
  `(1.5\, e^(t))` with respect to time.
• Find an expression for distance travelled by integrating speed with respect to time.
• Solve for the time
  `t` required to travel the given distance of
  `s = 20\; {\rm m}`.
• Substitute the value of
  `t` into the expression for speed and evaluate to find the speed at the given time.

To find an expression for speed
`v` at time
`t`, integrate the expression for the rate of change in speed with respect to time:

`\begin{aligned} v(t) &= \int (1.5\, e^(t))\, d t \\ &= 1.5\, \int e^(t)\, d t \\ &= 1.5\, e^(t) + C_(v)\end{aligned}`,

Where
`C_(v)` is a constant. Given that the vehicle started from rest,
`v(0) = 0`. Make use of this equality to find the value of
`C_(v)`:

`1.5\, e^(0) + C_(v) = v(0) = 0`.

`1.5 + C_(v) = 0`.

`C_(v) = -1.5`.

Hence, the expression for speed at time
`t` would be:

`v(t) = 1.5\, e^(t) - 1.5`.

To find an expression for distance
`s` travelled at time
`t`, integrate the expression for speed with respect to time:

`\begin{aligned} s(t) &= \int v(t) \, d t \\ &= \int \left(1.5\, e^(t) - 1.5\right) \, d t \\ &= 1.5\, e^(t) - 1.5\, t + C_(s)\end{aligned}`,

Where
`C_(s)` is also a constant. Assuming that the distance was initially
`0`,
`s(0) = 0`. Solve for the value of
`C_(s)`:

`1.5\, e^(0) - 1.5\, (0) + C_(s) = s(0) = 0`.

`1.5 + C_(s) = 0`

`C_(s) = -1.5`.

Hence, the expression for distance travelled at time
`t` would be:

`s(t) = 1.5\, e^(t) - 1.5\, t - 1.5`.

Assuming that
`t > 0`, solve this expression for the value of
`t` that would ensure
`s = 20`:

`1.5\, e^(t) - 1.5\, t - 1.5 = 20`.

`t \approx 2.8436` (
`t > 0`.)

Substitute the value of
`t` back into the expression for speed to obtain:

`v(t) \approx 1.5\, e^(2.8436) - 1.5 \approx 15.678`.

In other words, speed of the vehicle would be approximately
`15.678\; {\rm m\cdot s^(-1)}` when the distance travelled from the starting position is
`20\; {\rm m}`.