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3 years ago

Calculate the acceleration of a 1400-kg car that stops from 32 km/h "on a dime" (on a distance of 1.7 cm).

Physics

1 answer:

Wittaler [7]3 years ago

3 0

<h2>Answer:</h2>

-2324.4m/s²

<h2>Explanation:</h2>

Use one of the equations of motion as follows;

v² = u² + 2as ------------------------(i)

Where;

v = final velocity of object

u = initial velocity of object

a = acceleration of the object

s = distance covered by the object

From the question, the object is the car and it has the following;

v = final velocity of the car = 0 (since it comes to a stop)

u = initial velocity of the car = 32km/h = (32 x 1000/3600) m/s = 8.89m/s

a = unknown

s = on a dime = 1.7cm = 0.017m

Substitute these values into equation (i) as follows;

=> 0² = 8.89² + (2 x a x 0.017)

=> 0 = 79.03 + (0.034a)

=> 0.034a = 0 - 79.03

=> 0.034a = - 79.03

=> a = - $\frac{79.03}{0.034}$

=> a = - 2324.4m/s²

Therefore, the acceleration of the car is -2324.4m/s².

Note that the negative sign indicates that the car is actually decelerating.

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17. A 1350 g projectile is launched with a force of 150 N. The length of the firing arm is 1.25 m.

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a) The exit velocity of the projectile is 16.7 m/s

b) The maximum height achieved by the projectile is 14.2 m

c) The total time of flight is 3.40 s

d) The distance covered by the projectile is 46.5 m

Explanation:

We solve this first part of the problem by applying the work-energy theorem, which states that the work done on the projectile is equal to the gain in kinetic energy of the projectile. Mathematically:

$W=Fd = \frac{1}{2}mv^2-\frac{1}{2}mu^2=\Delta K$

where:

F = 150 N is the force applied

d = 1.25 m is the displacement of the projectile (the length of the firing arm)

m = 1350 g = 1.35 kg is the mass of the projectile

u = 0 is the initial velocity of the projectile

v is the exit velocity of the projectile

Solving for v, we find:

$v=\sqrt{\frac{2Fd}{m}}=\sqrt{\frac{2(150)(1.25)}{1.35}}=16.7 m/s$

Assuming the projectile is fired vertically upward, then the initial kinetic energy of the projectile as soon as he leaves the cannon is fully converted into gravitational potential energy as it reaches the top of its trajectory. So we can write:

$K_i = U_f$

$\frac{1}{2}mv^2=mgh$

where:

$K_i$ is the initial kinetic energy

$U_f$ is the final potential energy

m = 1350 g = 1.35 kg is the mass of the projectile

v = 16.7 m/s is the velocity at which the projectile leaves the cannon

$g=9.8 m/s^2$ is the acceleration of gravity

h is the maximum height reached by the projectile

And solving for h, we find

$h=\frac{v^2}{2g}=\frac{(16.7)^2}{2(9.8)}=14.2 m$

Assuming the projectile is launched vertically upward, then the total time of flight is twice the time it takes for reaching the maximum height. This time can be found by using the following suvat equation:

$v=u-gt$