Question

A car starts from rest and travels for 4.0s with a uniform acceleration of + 1.3m/s^2. The driver then applies the brakes, causing a uniform acceleration of - 1.8m/s^2. The breaks are applied for 1.50s.

a. How fast is the car going at the end of the braking period?
b. How far has the car gone from it’s start?

Thank you in advance!

Answer 1

V₂ = 2.5 m/s

S = 16 m

Step-by-step explanation:

Given:

V₀ = 0 m/s

t₁ = 4.0 s

a₁ = + 1.3 m/s²

a₂ = - 1.8m/s²

t₂ = 1.5 s

____________

V₂ - ?

S - ?

V₁ = V₀ + a₁·t₁ = 0 + 1.3·4.0 = 5.2 m/s

S₁ = (V₁² - V₀²) / (2·a₁) = ( 5.2² - 0²) / (2·1.3) = 10.4 m

V₂ = V₁ + a₂·t₂ = 5.2 + ( -1.8)·1.5 = 2.5 m/s

S₂ = (V₂² - V₁²) / (2·a₂) = ( 2.5² - 5.2²) / (2·(-1.8)) ≈ 5.8 m

S = S₁ + S₂ = 10.4 + 5.8 ≈ 16 m

Answer 2

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

Approximately
`16\; {\rm m}`.

Step-by-step explanation:

When acceleration is constant, the SUVAT equations would apply. Let
`a` denote acceleration,
`u` denote initial velocity,
`v` denote velocity after acceleration,
`x` denote displacement, and
`t` denote the duration of acceleration.

• 
  `v = u + a\, t` relates the velocity after acceleration to the duration of the acceleration.
• 
  `x = (v^(2) - u^(2)) / (2\, a)` relates the displacement after acceleration to initial velocity, final velocity, and acceleration.

Right before braking, the vehicle (initial velocity:
`u = 0\; {\rm m\cdot s^(-1)}`) would have accelerated at
`a = 1.3\; {\rm m\cdot s^(-2)}` for
`t = 4.0\; {\rm s}`. Apply the equation
`v = u + a\, t` to find the velocity of the vehicle after this period of acceleration:

`\begin{aligned} v &= u + a\, t \\ &= (0\; {\rm m\cdot s^(-1)}) + (1.3\; {\rm m\cdot s^(-2)})\, (4.0\; {\rm s}) \\ &= 5.2\; {\rm m\cdot s^(-1)}\end{aligned}`.

Apply the equation
`x = (v^(2) - u^(2)) / (2\, a)` to find the displacement
`x` of this vehicle at that moment:

`\begin{aligned} x &= (v^(2) - u^(2))/(2\, a) \\ &= \frac{(5.2\; {\rm m\cdot s^(-1)})^(2) - (0\; {\rm m\cdot s^(-1)})^(2)}{2 * 1.3\; {\rm m\cdot s^(-2)}} \\ &= 10.4\; {\rm m}\end{aligned}`.

It is given that acceleration is constant (at
`a = (-1.8\; {\rm m\cdot s^(-2)})`) during braking. The velocity before this period of acceleration (initial velocity) would now be
`u = 5.2\; {\rm m\cdot s^(-1)}`. After
`t = 1.50\; {\rm s}` of braking, the velocity of this vehicle would be:

`\begin{aligned} v &= u + a\, t \\ &= (5.2\; {\rm m\cdot s^(-1)}) + (-1.8\; {\rm m\cdot s^(-2)})\, (1.50\; {\rm s}) \\ &= 2.5\; {\rm m\cdot s^(-1)}\end{aligned}`.

Apply the equation
`x = (v^(2) - u^(2)) / (2\, a)` to find the displacement of this vehicle during this
`t = 1.50\; {\rm s}` of braking. Note that this displacement gives the position relative to where this vehicle started braking, not where it started from rest.

`\begin{aligned} x &= (v^(2) - u^(2))/(2\, a) \\ &= \frac{(2.5\; {\rm m\cdot s^(-1)})^(2) - (5.2\; {\rm m\cdot s^(-1)})^(2)}{2 * (-1.8)\; {\rm m\cdot s^(-2)}} \\ &= 5.775\; {\rm m}\end{aligned}`.

The total displacement of this vehicle (relative to where it started from rest) is the sum of the displacement during acceleration and the displacement during braking:
`10.4\; {\rm m} + 5.775\; {\rm m} \approx 16\; {\rm m}`. In other words, after that
`t = 1.50\; {\rm s}` of braking this vehicle would have been approximately
`16\; {\rm m}` from where it started.