Asked by Naomi
A car, moving along a straight stretch of highway, begins to accelerate at 0.0442 m/s^2. It takes the car 30.8 s to cover 1 km. How fast was the car going when it first began to accelerate?
Answer in units of m/s.
I tried converting 30.8 km/sec to m/s then subtracting 0.0442, but I got -0.0134, and it can't be a negative. I've also tried the answers 30799.558, 13.4, and 0.0134 but my online homework sheet has said they are all wrong. Thanks for the help!
Answer in units of m/s.
I tried converting 30.8 km/sec to m/s then subtracting 0.0442, but I got -0.0134, and it can't be a negative. I've also tried the answers 30799.558, 13.4, and 0.0134 but my online homework sheet has said they are all wrong. Thanks for the help!
Answers
Answered by
jzee11
Here what i did.
a=0.0442 m/s^2
t1=30.8 s
d=1 km =1000 m.
find the initial velocity (speed)=Vo when car starts accelerating.
From basic equation of acceleration,
a = dv/dt = (V1-Vo)/(t1-t0) ; t0= 0s.
Vo = V1 - a.t1
missing V1=d/t0 = 1000/30.8 = 32.46 m/s.
Then, Vo = 32.46 - (0.0442)(30.8)
=31.1 m/s
tell me answer is correct or not.
a=0.0442 m/s^2
t1=30.8 s
d=1 km =1000 m.
find the initial velocity (speed)=Vo when car starts accelerating.
From basic equation of acceleration,
a = dv/dt = (V1-Vo)/(t1-t0) ; t0= 0s.
Vo = V1 - a.t1
missing V1=d/t0 = 1000/30.8 = 32.46 m/s.
Then, Vo = 32.46 - (0.0442)(30.8)
=31.1 m/s
tell me answer is correct or not.
Answered by
jjd
jzee11 that answer wasn't right. I had the exact same numbers as the OP and it said that answer was wrong
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