CBSE Class 9 Answered
A car is traveling with a speed of 54 km/hr the driver applies the break and retards the car uniformly.the car is stopped in 5 seconds
1) The redardation of the car
2)Distance travelled before it is stopped after apply the break
Asked by rsaranya658 | 30 Jun, 2022, 02:44: PM
Initial speed = 54 km/hr = 54 × (5/18) = 15 m/s
Retardation a is determined from following equation
" v = u - ( a × t ) "
where v is final speed that is zero because after time t the car comes to rest. Initial speed is u
a = u / t = 15/5 = 3 m/s2
Distance S travelled is determined from the following equation
" S = ( u t ) - [ (1/2) a t2 ] "
S = ( 15 × 5 ) - [ (1/2) × 3 × 5 × 5 ]
S = 37.5 m
Answered by Thiyagarajan K | 30 Jun, 2022, 03:02: PM
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