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प्रश्न
Solve the following problem.
A car moving along a straight road with a speed of 120 km/hr, is brought to rest by applying brakes. The car covers a distance of 100 m before it stops. Calculate
(i) the average retardation of the car
(ii) time taken by the car to come to rest.
उत्तर
Given: u = 120 kmh-1 = `120 xx 5/18 = 100/3 "ms"^-1`, s = 100 m, v = 0.
To find:
1. Average retardation of the car (a)
2. Time taken by car (t)
Formulae:
1. v2 - u2 = 2as
2. v = u + at
Calculation:
From formula (i),
`0 - (100/3)^2 = "2a" xx 100`
∴ a = `(- 10000)/9 xx 1/200`
∴ a = `- 50/9`ms-2
From formula (ii),
t = `("v - u")/"a" = (0 - 100/3)/(- 50/9)` = 6 s
- Average retardation of the car is `underline(50/9 "ms"^-2)` (in magnitude).
- Time taken by the car to come to rest is 6 s.
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