A car runs 14 2/3
km using 1 litre of petrol. How much distance will it cover using
Answers
Answer:
Given:-
Car starts from rest at velocity 10m/s in 40 secs.
The driver applies a breakand slows down to 5m/s in 10 secs.
To Find:-
Acceleration in both the cases
Law used:-
{\bf{\boxed{First\:Law\:of\:Motion:\: v =u+at}}}
FirstLawofMotion:v=u+at
Here,
a = Acceleration
v = Final Velocity
u = Initial Velocity
t = time
Solution:-
Let the acceleration before applying breaks be a and after applying be a'.
Case 1: When the breaks were not applied.
Using, v = u + at
Here,
v = 10m/s
u = 0m/s
t = 40 sec
a = a
Putting Values,
\sf •\leadsto\:a =\dfrac{10-0}{40}•⇝a=
40
10−0
\sf •\leadsto\:a =\dfrac{10}{40}•⇝a=
40
10
\sf •\leadsto\:a =\dfrac{1}{4}•⇝a=
4
1
{\bf{\boxed{•\leadsto\:a= 0.25ms^{-2}}}}
•⇝a=0.25ms
−2
Hence, The acceleration of the car before applying breaks is 0.25m/s²
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Case 2: When breaks were applied.
Again Using, v = u + at
Here,
v = 5m/s
u = 10m/s
t = 10 sec
a = a'
Putting Values,
\sf •\leadsto\:a' =\dfrac{5-10}{10}•⇝a
′
=
10
5−10
\sf •\leadsto\:a' =\dfrac{-5}{10}•⇝a
′
=
10
−5
\sf •\leadsto\:a' =\dfrac{-1}{2}•⇝a
′
=
2
−1
{\bf{\boxed{•\leadsto\:a' = -0.5ms^{-2}}}}
•⇝a
′
=−0.5ms
−2
{negative sign show retardation}
Hence, The acceleration of the car after applying breaks is 0.5m/s².
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