Acceleration when time is not given

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To calculate the cyclist's acceleration when stopping, the correct approach involves using the formula V_f^2 = V_i^2 + 2ax, where V_f is the final velocity (0 m/s), V_i is the initial velocity (converted to m/s), and x is the distance (15 m). Initially, the cyclist's velocity was given in km/hr, which needed conversion to m/s for accurate calculations. The calculations revealed an error in the initial attempt, as the correct units were not used. After converting the velocity to SI units, the correct acceleration was determined to be -1.1 m/s². Proper unit conversion is crucial for accurate physics problem-solving.
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1. A cyclist traveling 21km/hr stops over a distance of 15m. His total mass is 73kg. What is his acceleration?



2. T=D/V, A=V/t



3. Since time is displacement over velocity, I can use the displacement and velocity to find time and then solve acceleration by a=v/t. Am I on the right track or should the mass have come into play there?
 
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V_f^2=V_i^2+2ax would work?
 
PhysicsAdvice said:
3. Since time is displacement over velocity, I can use the displacement and velocity to find time and then solve acceleration by a=v/t. Am I on the right track or should the mass have come into play there?

No-

x = vt

...is only valid when v is constant, which it is not.
 
I tried V_f^2=V_i^2+2ax which comes to
0=441+2a (225),
-441/225=2a,
-1.96=2a,
-0.98=a

but the answer is -1.1, error?
 
PhysicsAdvice said:
I tried V_f^2=V_i^2+2ax which comes to
0=441+2a (225),
-441/225=2a,
-1.96=2a,
-0.98=a

but the answer is -1.1, error?

I'm not sure what units you're using? Convert everything to SI units (velocity in m/s, distance in m) and it turns out fine.
 
vi=21km/hr vf=0km/hr distance=15m, so do i need to convert the 21km/hr to m/s?
 
nevermind I tried it again and found the right answer, thanks!
 
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