A car accelerates uniformly from 18km/hr to 36 km/hr in 5 sec.calculate the distance covered by the car in 5 sec.
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u=18 km/h=5 m/s
v=36 km/h=10 m/s
t=5 s
a=v-u/t
=10-5/5=5/5=1 m/s2
s=ut+1/2 at2
=5 (5) +1/2*1*5*5
=25+12.5=37.5m
v=36 km/h=10 m/s
t=5 s
a=v-u/t
=10-5/5=5/5=1 m/s2
s=ut+1/2 at2
=5 (5) +1/2*1*5*5
=25+12.5=37.5m
faizul:
wrong ans
What's the right answer
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Find acceleration of the car using the formula
A = (V-U)/T
where V is the final velocity, U is the initial velocity and T is time required.
But be careful since the velocities are in km/hr and time is in seconds.
18 km/hr = 5m/s.
36 km/ hr = 10m/s
Now we can find Acceleration of the car.
》A = (V-U)/T
》A = (10-5)/5
》A = 1 m/s^2
Now using the formula V^2 = U^2 + 2×A×S find S, which is the distance covered.
》10^2 = 5^2 + 2×1×S
》100 = 25 + 2×S
》2×S = 75
》S = 37.5 m
Hence the car covers a distance of 37.5 m according to the given data
A = (V-U)/T
where V is the final velocity, U is the initial velocity and T is time required.
But be careful since the velocities are in km/hr and time is in seconds.
18 km/hr = 5m/s.
36 km/ hr = 10m/s
Now we can find Acceleration of the car.
》A = (V-U)/T
》A = (10-5)/5
》A = 1 m/s^2
Now using the formula V^2 = U^2 + 2×A×S find S, which is the distance covered.
》10^2 = 5^2 + 2×1×S
》100 = 25 + 2×S
》2×S = 75
》S = 37.5 m
Hence the car covers a distance of 37.5 m according to the given data
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