A cyclist is traveling at 15 m/s she applies breaks so as to just avoid collision with a wall 18 m away . What declaration must be she have ? also calculate time taken during this process ?
Answers
Given:-
- Initial Velocity = 15m/s
- Distance of wall from him = 18m
- Final Velocity = 0m/s( As he applied brakes).
To Find:-
- Declaration of the cycle and time taken during thus Process.
Formulae used:-
- v² - u² = 2as
- v = u + at
where,
v = Final Velocity
u = Initial Velocity
a = Acceleration
S = Distance
t = Time.
Now,
v² - u² = 2as
(0)² - (15)² = 2 × a × s
0 - 225 = 2 × 18 × a
-225 = 36a
a = -225/36
a = -6.25m/s²
Hence, The retardation is 6.25m/s²
Therefore,
Using first equation to find Time taken
v = u + at
0 = 15 + 6.25 × t
-15 = 6.25t
t = -15/6.25
t = 2.4s.
Hence, The time taken During this Process is 2.4 Secoond.
Explanation:
Given; initial velocity is 15 m/s, distance covered is 18 m and final velocity is 0 m/s (as brakes are applied).
To find: deceleration and time.
Using the third equation of motion,
v² - u² = 2as
Substitute the values,
→ (0)² - (15)² = 2a(18)
→ -225 = 36a
→ -6.25 = a
Hence, the retardation of the cyclist is 6.25 m/s².
Now,
Using the first equation of motion,
v = u + at
Substitute the values,
→ 0 = 15 + (-6.25)t
→ -15 = -6.25t
→ 2.4 = t
Hence, the time taken by the cyclist is 2.4 seconds.