Answer:
Answer:
2.5\; {\rm m\cdot s^{-1}}2.5m⋅s
−1
.
Approximately 16\; {\rm m}16m .
Step-by-step explanation:
When acceleration is constant, the SUVAT equations would apply. Let aa denote acceleration, uu denote initial velocity, vv denote velocity after acceleration, xx denote displacement, and tt denote the duration of acceleration.
v = u + a\, tv=u+at relates the velocity after acceleration to the duration of the acceleration.
x = (v^{2} - u^{2}) / (2\, a)x=(v
2
−u
2
)/(2a) 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}}u=0m⋅s
−1
) would have accelerated at a = 1.3\; {\rm m\cdot s^{-2}}a=1.3m⋅s
−2
for t = 4.0\; {\rm s}t=4.0s . Apply the equation v = u + a\, tv=u+at to find the velocity of the vehicle after this period of acceleration:
\begin{gathered}\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}\end{gathered}
v
=u+at
=(0m⋅s
−1
)+(1.3m⋅s
−2
)(4.0s)
=5.2m⋅s
−1
.
Apply the equation x = (v^{2} - u^{2}) / (2\, a)x=(v
2
−u
2
)/(2a) to find the displacement xx of this vehicle at that moment:
\begin{gathered}\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a} \\ &= \frac{(5.2\; {\rm m\cdot s^{-1}})^{2} - (0\; {\rm m\cdot s^{-1}})^{2}}{2 \times 1.3\; {\rm m\cdot s^{-2}}} \\ &= 10.4\; {\rm m}\end{aligned}\end{gathered}
x
=
2a
v
2
−u
2
=
2×1.3m⋅s
−2
(5.2m⋅s
−1
)
2
−(0m⋅s
−1
)
2
=10.4m
.
It is given that acceleration is constant (at a = (-1.8\; {\rm m\cdot s^{-2}})a=(−1.8m⋅s
−2
) ) during braking. The velocity before this period of acceleration (initial velocity) would now be u = 5.2\; {\rm m\cdot s^{-1}}u=5.2m⋅s
−1
. After t = 1.50\; {\rm s}t=1.50s of braking, the velocity of this vehicle would be:
\begin{gathered}\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}\end{gathered}
v
=u+at
=(5.2m⋅s
−1
)+(−1.8m⋅s
−2
)(1.50s)
=2.5m⋅s
−1
.
Apply the equation x = (v^{2} - u^{2}) / (2\, a)x=(v
2
−u
2
)/(2a) to find the displacement of this vehicle during this t = 1.50\; {\rm s}t=1.50s of braking. Note that this displacement gives the position relative to where this vehicle started braking, not where it started from rest.
\begin{gathered}\begin{aligned} x &= \frac{v^{2} - u^{2}}{2\, a} \\ &= \frac{(2.5\; {\rm m\cdot s^{-1}})^{2} - (5.2\; {\rm m\cdot s^{-1}})^{2}}{2 \times (-1.8)\; {\rm m\cdot s^{-2}}} \\ &= 5.775\; {\rm m}\end{aligned}\end{gathered}
x
=
2a
v
2
−u
2
=
2×(−1.8)m⋅s
−2
(2.5m⋅s
−1
)
2
−(5.2m⋅s
−1
)
2
=5.775m
.
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}10.4m+5.775m≈16m . In other words, after that t = 1.50\; {\rm s}t=1.50s of braking this vehicle would have been approximately 16\; {\rm m}16m from where it started.