Question

a car can travel a maximum speed of 180 km/h and can have maximum acceleration 5m/s^2 and retardation of 3m/s^2. How fast can it start from rest and come to rest in travelling 300m​

Answer 1

Answer:8 root 5 sec

Explanation:

Answer 2

Answer:

The answer is 26.67 seconds.

Explanation:

Given values are -

• The initial velocity (u) is 0 m/s.
• The maximum acceleration (a) is 5 m/s².
• The maximum speed till it starts retarding (v) is 180 km/h, which is 50 m/s.
• The distance traveled by car is A from u to v and the time taken is t₁.
• Then the car starts retarding with retardation (r) of 3 m/s².
• The final speed (V) where it stops is 0 m/s.
• The distance traveled by car is B from v to V and the time taken is t₂.
• The total distance the car traveled (A+B) is 300 meters.

1) From u to v

v² - u² = 2as

(50)² - (0)² = 2(5)(A)

2500 = 10(A)

A is 250 meters

2). B = 300 - A = 300 - 250 = 50 meters

3) s = ut + (½)at²

250 = 0(t₁) + (½)(5)(t₁)²   [s is A here, 250 meters)

t₁= 10 s

4). v = u + at

0 = 50 + (-3)(t₂)      [v is V here, 0 m/s and a is r here which is -3m/s²]

t₂ = 16.67 s

Total time taken by the car to travel 300 meters distance is (t₁+t₂) = 10 + 16.67 = 26.67 s

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