The driver of a car travelling along a straight road with a speed of 72 km/hr observes a signboard which gives the speed limit to be 54 km/hour. The signboard is 70 m ahead when the driver applies the brakes. Calculate the acceleration of the car which will cause the car to pass the signboard at the stated speed limit.
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Answered by
239
u = 72 km/h = 20 m/s
v = 54 km/h = 15 m/s
S = 70 m
a = ?
v² = u² - 2×a×S (because there will be deceleration)
15² = 20² - 2×a×70
225 = 400 - 140 a
140 a = 400-225
140 a = 175
a = 175/140
a = 1.25 m/s²
Acceleration will be -1.25 m/s².
v = 54 km/h = 15 m/s
S = 70 m
a = ?
v² = u² - 2×a×S (because there will be deceleration)
15² = 20² - 2×a×70
225 = 400 - 140 a
140 a = 400-225
140 a = 175
a = 175/140
a = 1.25 m/s²
Acceleration will be -1.25 m/s².
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Answered by
83
The kinematic equation is simply V² - U² = 2 a S
V = 54 kmph = 15 m/s U = 72 kmph = 72*5/18 = 20 m/s
a = (15² - 20² ) / 2 * 70 = - 175 / 140 =- 5/4 = - 1.25 meters /sec
the car has 5/1.25 = 4 seconds to pass the signboard at the specfied speed
V = 54 kmph = 15 m/s U = 72 kmph = 72*5/18 = 20 m/s
a = (15² - 20² ) / 2 * 70 = - 175 / 140 =- 5/4 = - 1.25 meters /sec
the car has 5/1.25 = 4 seconds to pass the signboard at the specfied speed
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