A car moving with an initial velocity u is brought to rest by application of brakes which provides uniform retardation of 2.5ms–2 for 10s. The value of u is:
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Explanation:
nitial velocity, u=?
Final velocity, v=0m/s (car is stopped)
Acceleration, a=−2.5m/s
2
(negative because it is retardation)
Time, t=10s
Use the relation of the first equation of motion:
v=u+at
0=u+(−2.5)×10
u=25m/s
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When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity (v
0
) and the braking capacity, or deceleration, -a that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of v
0
and a.
Velocity of moving vehicle=v
0
and decceleration =a
Final velocity of the vehicale =0 (As it stops)
Let the stopping distance be s
Then applying the third equation of motion we have:
v
2
=v
0
2
+2(−a)s
⇒0=v
0
2
−2as
⇒s=
2a
v
0
2
→ Expression for stopping distance.
0
) and the braking capacity, or deceleration, -a that is caused by the braking. Derive an expression for stopping distance of a vehicle in terms of v
0
and a.
Velocity of moving vehicle=v
0
and decceleration =a
Final velocity of the vehicale =0 (As it stops)
Let the stopping distance be s
Then applying the third equation of motion we have:
v
2
=v
0
2
+2(−a)s
⇒0=v
0
2
−2as
⇒s=
2a
v
0
2
→ Expression for stopping distance.
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