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Y_Kistochka [10]
4 years ago
8

A car is initially moving at 35 km/h along a straight highway. To pass another car, it speeds up to 135 km/h in 10.5 seconds at

a constant acceleration.
(a) how large was the acceleration in m/s ^2
(b)how large was the acceleration, in units go g= 9.80 m/s ^2
Physics
1 answer:
Aleks04 [339]4 years ago
6 0
Acceleration = (velocity final-velocity initial)/ time
where
velocity final = 135 km/hr x 1 hr /3600 s x 1000m/1km
                     = 37.5 m/s
velocity initial = 35 km/hr x  1hr /3600 s x 1000 m/1 km
                      =  9.72 m/s
a) acceleration = 2.646 m/s^2
b) acceleration in g units  = (2.646m/s^2)/(9.8m/s^2)
                                              = 0.27 units

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Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m
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Answer: m \frac{d}{dt}v_{(t)}

Explanation:

In the image  attached with this answer are shown the given options from which only one is correct.

The correct expression is:

m \frac{d}{dt}v_{(t)}

Because, if we derive velocity v_{t} with respect to time t we will have acceleration a, hence:

m \frac{d}{dt}v_{(t)}=m.a

Where m is the mass with units of kilograms (kg) and a with units of meter per square seconds \frac{m}{s}^{2}, having as a result kg\frac{m}{s}^{2}

The other expressions are incorrect, let’s prove it:

\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)} This result has units of kg\frac{m}{s}

m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m This result has units of kg

m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C This result has units of kgm^{2} and C is a constant

m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m This result has units of kg

m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m This result has units of kg

\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C This result has units of kg \frac{m^{3}}{s^{3}} and C is a constant

m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C This result has units of kg \frac{m^{2}}{s^{4}} and C is a constant

\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0 because v_{(x)} is a constant in this derivation respect to t

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6 0
3 years ago
A car is traveling at a speed of 50 m/sec when it suddenly accelerates at a rate of 5 m/sec2. How fast will the car be going (fi
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Answer:

200 ms-1

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5 0
3 years ago
a lorry travels 3600m on a test track accelerating constantly at 3m/s squared from standstill. what is the final velocity (3 sig
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The final velocity of the truck is found as 146.969 m/s.

Explanation:

As it is stated that the lorry was in standstill position before travelling a distance or covering a distance of 3600 m, the initial velocity is considered as zero. Then, it is stated that the lorry travels with constant acceleration. So we can use the equations of motion to determine the final velocity of the lorry when it reaches 3600 m distance.

Thus, a initial velocity (u) = 0, acceleration a = 3 m/s² and the displacement s is 3600 m. The third equation of motion should be used to determine the final velocity as below.

2as =v^{2} - u^{2} \\\\v^{2} = 2as + u^{2}

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Thus,  the final velocity of the truck is found as 146.969 m/s.

8 0
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