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3 years ago

a car moving east at 45 km/h turns and travels west at 30 km/h. What is the magnitude and direction of the change in velocity?

Physics

2 answers:

slega [8]3 years ago

7 0

Answer:

The magnitude and direction of the change in velocity is 75 m/s in west.

Explanation:

Given that,

Velocity of car = 45 km/h in east

Velocity of car = 30 km/h in west

We need to calculate the magnitude of the change in velocity

Using formula of change in velocity

Change in velocity = final velocity - initial velocity

$\Delta v=V_{f}-v_{i}$

Put the value into the formula

$\Delta v=-30-45$

$|\Delta v|=75\ m/s$

The direction of the the change in velocity is toward west.

Hence, The magnitude and direction of the change in velocity is 75 m/s in west.

PSYCHO15rus [73]3 years ago

3 0

Magnitude of change is 15km/h
change in direction is to the right?

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During a very quick stop, a car decelerates at 6.8 m/s^2. Assume the forward motion of the car corresponds to a positive directi

geniusboy [140]

Answer:

-24.28571 rad/s²

29.57239 revolutions

3.91176 seconds

52.026478 m

Explanation:

$a_t$ = Tangential acceleration = -6.8 m/s²

r = Radius of wheel = 0.28

$\omega_i$ = Initial angular velocity = 95 rad/s

$\theta$ = Angle of rotation

$\omega_f$ = Final angular velocity

t = Time taken

Angular acceleration is given by

$\alpha=\frac{a_t}{t}\\\Rightarrow \alpha=\frac{-6.8}{0.28}\\\Rightarrow \alpha=-24.28571\ rad/s^2$

The angular acceleration is -24.28571 rad/s²

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-95^2}{2\times -24.28571}\\\Rightarrow \theta=185.80885\ rad=185.80885\times \frac{1}{2\pi}\\\Rightarrow \theta=29.57239\ rev$