All categories

3 years ago

A vehicle moving along at 5m/s. What should be the constant deceleration in order to stop it within 15m?

Physics

1 answer:

Oduvanchick [21]3 years ago

5 0

Answer:

a = -5/6 m/s²

Explanation:

v² = u² + 2as

a = (v² - u²) / 2s

a = (0² - 5²) / 2(15)

a = -5/6 m/s²

This assumes the initial velocity was in the positive direction.

You might be interested in

A 6cm diameter horizontal pipe gradually narrows to 4cm.

Rom4ik [11]

Answer: $Q=0.5612 m^3/s$

Explanation:

Given

diameter of pipe( $d_1$ )=6 cm

diameter of pipe( $d_2$ )=4 cm

$P_1=32 kPa$

$P_2=24 kPa$

$A_1=\frac{\pi }{4}6^2=9\pi cm^2$

$A_2=\frac{\pi }{4}4^2=4\pi cm^2$

$v_1=\frac{Q}{A_1}$

Applying bernoulli's equation

$\frac{P_1}{\rho g}+\frac{v^2_1}{2g}+z_1=\frac{P_2}{\rho g}+\frac{v^2_2}{2g}+z_2$

$\frac{P_1}{\rho g}+\frac{\frac{Q^2}{A_1^2}}{2g}+z_1=\frac{P_2}{\rho g}+\frac{\frac{Q^2}{A_2^2}}{2g}+z_2$

since $z_1=z_2$

$\frac{32\times 10^3}{10^3\times 9.81}+\frac{Q^2}{2A_1^2g}=\frac{24\times 10^3}{10^3\times 9.81}+\frac{Q^2}{2A_2^2g}$

$Q^2=\frac{8\times 2\times 81\pi ^2\times 16\pi ^2\times 10^{-4}}{65\pi ^2}$

$Q^2=3149.3722\times 10^{-4}$

$Q=\sqrt{3149.3722\times 10^{-4}}$

$Q=0.5612 m^3/s$

4 0

3 years ago

Starting from rest, a runner reaches a speed of 2.8 m/s in 2.1 s. In the same time 2.1 s time, a motorcycles increases speed fro

maw [93]

Answer:

a) Acceleration of runner is 1.33 m/s²

b) Acceleration of motorcycle is 2.85 m/s²

c) The motorcycle moves 84.21-2.94 = 81.06 m farther than the runner.

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement

a = Acceleration

Equation of motion

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{2.8-0}{2.1}\\\Rightarrow a=1.33\ m/s^2$

Acceleration of runner is 1.33 m/s²

$v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{43-37}{2.1}\\\Rightarrow a=2.85\ m/s^2$

Acceleration of motorcycle is 2.85 m/s²

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{2.8^2-0^2}{2\times 1.33}\\\Rightarrow s=2.94\ m$

The runner moves 2.94 m

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{43^2-37^2}{2\times 2.85}\\\Rightarrow s=84.21\ m$

The motorcycle moves 84.21 m

The motorcycle moves 84.21-2.94 = 81.06 m farther than the runner.

3 0

3 years ago

An instrument is defective for all of the following conditions EXCEPT:_______

NARA [144]

Answer:

I believe its C

6 0

4 years ago

A jet, sitting on the runway, takes off and accelerates at 8.0 m/s for 16s. How far did the jet travel down the runway?

nydimaria [60]

Answer:

2.4 m/s". 1

Explanation:

A jet with mass m = 8 x 10* kg jet accelerates down the runway for takeoff at 2.4 m/s". 1

8 0

3 years ago

If you and your friend EACH apply 55N of force pushing against a wall for 10 minutes. How much work

alina1380 [7]

Answer: 0 J

Explanation: there’s no movement if it’s a wall

8 0

3 years ago

A vehicle moving along at 5m/s. What should be the constant deceleration in order to stop it within 15m?​

A vehicle moving along at 5m/s. What should be the constant deceleration in order to stop it within 15m?