The driver of a car travelling along a straight road with a speed of 72Km/h observes a signboard which gives the speed limit to be 54Km/h. The signboard is 70m 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?
Answers
Answered by
31
Initial speed is 72kmph = 20ms
Final is 54 kmph = 15ms
Distance is 0.07
v2-u2=2as
400-225= 2a(70)
a= - 35/28 = 1.25ms2
Final is 54 kmph = 15ms
Distance is 0.07
v2-u2=2as
400-225= 2a(70)
a= - 35/28 = 1.25ms2
bshivam466:
u=72* 5/18 = 20 m/s
Answered by
32
v= final velocity= 54 km/h= 54x 5/18= 15 m/s
u= initial velocity=72 km/h= 72x5/18= 20 m/s
s= distance = 70m
a= acceleration = ?
using the third equation of motion,
v²= u² + 2as,
we get,
15²= 20² + 2 x a x 70
225= 400 + 140a
225-400= 140a
140a = -175
a= -175/140 = -1.25 m/s²
∴ the acceleration of the car which will allow it to pass the signboard at 15m/s (54 km/h) is -1.25 m/s².
u= initial velocity=72 km/h= 72x5/18= 20 m/s
s= distance = 70m
a= acceleration = ?
using the third equation of motion,
v²= u² + 2as,
we get,
15²= 20² + 2 x a x 70
225= 400 + 140a
225-400= 140a
140a = -175
a= -175/140 = -1.25 m/s²
∴ the acceleration of the car which will allow it to pass the signboard at 15m/s (54 km/h) is -1.25 m/s².
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