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In a road test,car A accelerates from rest (0km/h)to 100.0 km/h in 16.0 and car b takes 8.0s in the same test.Which car has the greater average acceleration?By how many times?
show me the work please.

2007-01-30 14:34:38 · 4 answers · asked by Anonymous in Science & Mathematics Physics

4 answers

Alright, start with car A.
Divide: 100/16=6.25 which is 6.25 kilometers per second

Car B.
Divide 100/8=12.5 which is 12.5 kilometers per second

Therefore car B has the greater average acceleration by two times, or twice the acceleration.

2007-01-30 14:47:51 · answer #1 · answered by super_kick_chick 2 · 0 0

Assuming you mean 16.0 seconds:
Car B has twice the acceleration. 16.0/8.0=2
ok. ok
car A accelerates to 100 km/hr is 16 seconds.
Its acceleration is 100/16 =6.25 km/hr/sec
car B accelerates to 100 km/hr is 8 seconds.
Its acceleration is 100/8 =12.5 km/hr/sec
12.5>6.25 so car B has greater average acceleration.
12.5/6.25=2 times the acceleration.

2007-01-30 22:44:54 · answer #2 · answered by J C 5 · 0 0

v = u + at (u= initial velocity= 0), (1 hour = 3600 s)

for car A 100/3600 = 0 + a multiplied by 16

a = 100 / 3600 multiplied by 16

car B 100 / 3600 = 0 +a muliplied by 8

a = 100 / 3600 muliplied by 8


hence car B has greater av acceleration

car B has twice the acceleration of A

2007-01-30 22:50:27 · answer #3 · answered by calmserene 4 · 0 0

yes

2007-01-30 22:41:06 · answer #4 · answered by Anonymous · 0 0

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