A car is initially going 50 ft/sec brakes at a constant rate (constant negative acceleration), coming to a stop in 5 seconds.a) graph the velocity from&nbsp;t=0&nbsp;to&nbsp;t=5b) how far does the car travel?c) how far does the car travel if its initial velocity isdoubled, but it brakes at the same constant rate?

Initial velocity is u(t)=50ft/sec2 acceleration is =−a ft/sec then by the equation V(t)=u(t)−at v(t)=50−at when t=5 sec v(t)=0 then 0=50−5a a=10 Then the equation of the notion is v(t)=50−10t (b) distance traveled in 5 seconds is = area of the triangle =12⋅5⋅50=125 feet c) if the initial velocity is u(t)=100ft/sec then v(t)=100−at so v(t)=100−10t v(t)=0 when t=10 then the distance travelled in 10 seconds is = area of the triangle with base 10 and height 100 =12⋅100⋅10=500 feet

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Force, Motion and Energy
Acceleration Due to Gravity

geduiwelh

Answered question

2021-01-02

A car is initially going 50 ft/sec brakes at a constant rate (constant negative acceleration), coming to a stop in 5 seconds.
a) graph the velocity from $t = 0$ to $t = 5$
b) how far does the car travel?
c) how far does the car travel if its initial velocity isdoubled, but it brakes at the same constant rate?

Answer & Explanation

Neelam Wainwright

Skilled2021-01-03Added 102 answers

Initial velocity is $u (t) = 50 f t / \sec^{2}$
acceleration is $= - a f t / s e c$ then by the equation
$V (t) = u (t) - a t v (t) = 50 - a t$
when $t = 5 s e c v (t) = 0$
then $0 = 50 - 5 a$
$a = 10$
Then the equation of the notion is
$v (t) = 50 - 10 t$

(b) distance traveled in 5 seconds is
= area of the triangle $= \frac{1}{2} \cdot 5 \cdot 50 = 125$ feet
c) if the initial velocity is $u (t) = 100 f t / s e c$
then $v (t) = 100 - a t$
so $v (t) = 100 - 10 t$
$v (t) = 0$ when $t = 10$
then the distance travelled in 10 seconds is
= area of the triangle with base 10 and height 100
$= \frac{1}{2} \cdot 100 \cdot 10 = 500$ feet

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