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Talja [164]
3 years ago
10

In 1991 four English teenagers built an eletric car that could attain a speed of 30.0m/s. Suppose it takes 8.0s for this car to

accelerate from 18.0m/s to 30.0m/s. What is the magnitude of the car's acceleration?
Physics
2 answers:
ehidna [41]3 years ago
8 0

Answer:

Acceleration of the car, a=1.5\ m/s^2

Explanation:

It is given that,

Initial speed of the car, u = 18 m/s

Final speed of the car, v = 30 m/s

Time taken, t = 8 s

We need to find the acceleration of the car. Let it is given by a. It can be calculated using the first equation of motion.

v=u+at

Where

a is the acceleration of the car.

a=\dfrac{v-u}{t}

a=\dfrac{30-18}{8}      

a=1.5\ m/s^2

So, the magnitude of the car's acceleration is 1.5\ m/s^2. Hence, this is the required solution.                                          

ivanzaharov [21]3 years ago
6 0

a=Δv/Δt=(30.0-18.0)/8.0=12.0/8.0=1.5 m/s²


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Can anyone solve these for my by using unit vectors? Can you also please show your work
Oxana [17]

4. The Coyote has an initial position vector of \vec r_0=(15.5\,\mathrm m)\,\vec\jmath.

4a. The Coyote has an initial velocity vector of \vec v_0=\left(3.5\,\frac{\mathrm m}{\mathrm s}\right)\,\vec\imath. His position at time t is given by the vector

\vec r=\vec r_0+\vec v_0t+\dfrac12\vec at^2

where \vec a is the Coyote's acceleration vector at time t. He experiences acceleration only in the downward direction because of gravity, and in particular \vec a=-g\,\vec\jmath where g=9.80\,\frac{\mathrm m}{\mathrm s^2}. Splitting up the position vector into components, we have \vec r=r_x\,\vec\imath+r_y\,\vec\jmath with

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)t

r_y=15.5\,\mathrm m-\dfrac g2t^2

The Coyote hits the ground when r_y=0:

15.5\,\mathrm m-\dfrac g2t^2=0\implies t=1.8\,\mathrm s

4b. Here we evaluate r_x at the time found in (4a).

r_x=\left(3.5\,\dfrac{\mathrm m}{\mathrm s}\right)(1.8\,\mathrm s)=6.3\,\mathrm m

5. The shell has initial position vector \vec r_0=(1.52\,\mathrm m)\,\vec\jmath, and we're told that after some time the bullet (now separated from the shell) has a position of \vec r=(3500\,\mathrm m)\,\vec\imath.

5a. The vertical component of the shell's position vector is

r_y=1.52\,\mathrm m-\dfrac g2t^2

We find the shell hits the ground at

1.52\,\mathrm m-\dfrac g2t^2=0\implies t=0.56\,\mathrm s

5b. The horizontal component of the bullet's position vector is

r_x=v_0t

where v_0 is the muzzle velocity of the bullet. It traveled 3500 m in the time it took the shell to fall to the ground, so we can solve for v_0:

3500\,\mathrm m=v_0(0.56\,\mathrm s)\implies v_0=6300\,\dfrac{\mathrm m}{\mathrm s}

5 0
3 years ago
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