a train travelling at 72km/h undergoes a uniform retatation of 2m/s² when brakes are applied.find the time taken to come to rest and the distance travelled from the place where the brakes were applied
Answers
Answer:
Let's first convert km/h to m/s:
v_0 = 72 \dfrac{km}{h} \cdot \dfrac{1000 m}{1 km} \cdot \dfrac{1 h}{3600 s} = 20 \dfrac{m}{s}.v
0
=72
h
km
⋅
1km
1000m
⋅
3600s
1h
=20
s
m
.
We can find the distance traveled by the train from the place where the brakes were applied from the kinematic equation:
v_f^2 = v_0^2 + 2ad,v
f
2
=v
0
2
+2ad,
here, v_0 = 20 m/sv
0
=20m/s is the initial velocity of the train, v_f = 0v
f
=0 is the final velocity of the train, a = -2 m/s^2a=−2m/s
2
is the deceleration of the train, dd is the distance traveled by the train from the place where the brakes were applied.
Then, we get:
d = \dfrac{v_f^2 - v_0^2}{2a},d=
2a
v
f
2
−v
0
2
,
d = \dfrac{(0 \dfrac{m}{s})^2 - (20 \dfrac{m}{s})^2}{2 \cdot (-2 \dfrac{m}{s^2})} = 100 m.d=
2⋅(−2
s
2
m
)
(0
s
m
)
2
−(20
s
m
)
2
=100m.
Answer:
d = 100 m.d=100m.