A car initially at rest picks up a velocity of 72km/hr in 20s . If amass of car is 1000kg, calculate force developed by its engine and the distance covered by the car in 20s.
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Answered by
109
u = 0. v = 72 kmph = 72*5/18 = 20m/s.
t = 20 s. m = 1000 kg
a = (v-u)/t = 1 m/s^2
Force F = m a = 1000 N
OR... F = change in momentum/ t.
F = ( 1000*20-0)/20 = 1 kN
Distance traveled s = t *(v+u)/2 = 20 * 20/2 = 200m.
OR. .. s = u t + 1/2* a t ^2 = 20^2/2 = 200m.
t = 20 s. m = 1000 kg
a = (v-u)/t = 1 m/s^2
Force F = m a = 1000 N
OR... F = change in momentum/ t.
F = ( 1000*20-0)/20 = 1 kN
Distance traveled s = t *(v+u)/2 = 20 * 20/2 = 200m.
OR. .. s = u t + 1/2* a t ^2 = 20^2/2 = 200m.
kvnmurty:
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AionAbhishek, SamSid1: your formula for distance calculation was wrong. I asked for correction & I gave the reason too. But you didn't correct it.
Answered by
52
Hey there !!!!!!!!
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Mass of the car = 1000 kg
Initially velocity of car (u) = 0 Final Velocity = 72 kmph Time taken = 20s
Final velocity (v) = 72kmph = 72*5/18 = 4*5= 20m/s
v= u + at
v-u/t = a
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Force developed by engine (F)= ma = m(v-u)/t = 1000(20-0)/20 = 1000N
Distance traveled s=ut+at²/2 = 0+400/2 = 200 m
So force provided = 100N and Distance covered = 200m
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Hope this helped you...............
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Mass of the car = 1000 kg
Initially velocity of car (u) = 0 Final Velocity = 72 kmph Time taken = 20s
Final velocity (v) = 72kmph = 72*5/18 = 4*5= 20m/s
v= u + at
v-u/t = a
~~~~~~~~~~~~~~~~~~~~~~~~~~
Force developed by engine (F)= ma = m(v-u)/t = 1000(20-0)/20 = 1000N
Distance traveled s=ut+at²/2 = 0+400/2 = 200 m
So force provided = 100N and Distance covered = 200m
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Hope this helped you...............
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