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Harrizon [31]
3 years ago
11

a certain car is capable of accelerating at a rate of 10.60m/s^2. how long does it take for this car to go from a speed of 55/mi

/h to a speed of 60 mi/hr
Physics
1 answer:
irina [24]3 years ago
7 0
Use the equation:
v = u + at
because it has all the variables given and need to find. First you must make sure units are the same. Acceleration is given in m/s^2. Check to make sure this is correct before using the following solution. Convert acceleration from meters per second squared to miles per hour squared.

look up conversion meters to miles
1 mile = 1609.34 meters
10.60 m/s^2 * 1mile/1609.34m = 0.0069 mi/s^2
there are 3600 seconds in an hour, then there are 3600s*3600s in 1 hr^2
0.0069 mi/s^2 * 3600s*3600s/hr^2 = 85362 mi/hr^2

Then using: v = u +at
60 = 55 + 85362t
60 - 55 = 85362t
5 = 85362t
5/85362 = t
5.9 x 10^-5 hr = t
very small value in hours, multiply by 3600 to convert to seconds.

5.9 x 10^-5 hr * 3600s/hr = 0.21 seconds
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2.1) (i) W = mg downwards
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
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2.2.1)
v = 25 m/s
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∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s
a \:  =  \:  \frac{dv}{dt}  \:  =  \:  \frac{25 \:  \frac{m}{s} }{50 \: s} \:  =  \: 0.5 \:  \frac{m}{ {s}^{2} }
a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)
2ax \:  =  \:  {v}^{2}  \:  -  \:  {u}^{2}
x \:  =  \:  \frac{ {v}^{2}  \:   -  \:  {u}^{2} }{2a}  \:  =  \:   \frac{{(25 \:  \frac{m}{s})}^{2}  \:  -  \:  {(0 \:  \frac{m}{s} )}^{2} }{2 \:  \times  \: 0.5 \:  \frac{m}{ {s}^{2} } } \:  =  \: 625 \: m
2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.
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