An alert driver can apply the brakes fully in about 0.5 seconds. How far would the car travel if it

Physics

1 answer:

dybincka [34]4 years ago

8 0

Answer:

The car would travel after applying brakes is, d = 14.53 m

Explanation:

Given that,

The time taken to apply brakes fully is, t = 0.5 s

The velocity of the car, v = 29.06 m/s

The distance traveled by the car in 0.5 s, d = ?

The relation between the velocity, displacement, and time is given by the formula

d = v x t m

Substituting the values in the above equation,

d = 29.06 m/s x 0.5 s

= 14.53 m

Therefore, the car would travel after applying brakes is, d = 14.53 m

You might be interested in

In gas chromatography, what are the advantages of (a) temperature programming? (check all that apply.)

TiliK225 [7]

The following statements apply:
1. Resolution of low boiling solutes is maintained.
2. Retention times of high boiling solutes are decreased.
Temperature programming refers to the process of increasing the temperature of gas chromatography column as a function of time. Temperature programming is usually applied to samples which contain a mixture of components whose boiling points are within narrow ranges

3 0

3 years ago

Which equation represents a neutralization reaction?

JulijaS [17]

Option B: acid+base->water+salt is the equation represents a neutralization reaction.

6 0

2 years ago

What is the speed of a wave frequency of 300hz and a wavelength of 25M

tia_tia [17]

Answer:

7500 m/s

Explanation:

We can use the equation velocity of a wave equals wavelength times frequency. Therefore, v = wavelength*f = (25 m)(300 Hz) = m/s7,500

8 0

4 years ago

A car initially traveling at 27.7 m/s undergoes a constant negative acceleration of magnitude 2.00 m/s2 after its brakes are app

lys-0071 [83]

The car's velocity at time t is given by

$v=27.7\dfrac{\rm m}{\rm s}+\left(-2.00\dfrac{\rm m}{\mathrm s^2}\right)t$

It comes to a stop when v = 0, which happens when

$0=27.7\dfrac{\rm m}{\rm s}+\left(-2.00\dfrac{\rm m}{\mathrm s^2}\right)t\implies t=13.85\,\mathrm s$

or after about 13.9 s.

In this time, the car travels a distance x given by

$x=\left(27.7\dfrac{\rm m}{\mathrm s}\right)(13.85\,\mathrm s)+\dfrac12\left(-2.00\dfrac{\rm m}{\mathrm s^2}\right)(13.85\,\mathrm s)^2=191.823\,\mathrm m$