Question

the driver of a car slams on the brakes when he sees a tree blocking the road. the car slows significantly with an acceleration of -5.60 m/s^2 for 4.20 s, making straight skid marks 62.4 m long all the way to the tree. with what speed does the car then strike the tree?

@mathslover i got 10 something but my book says 3 something...how? o.O

Answer 1

the book is correct @lgbasallote

Answer 2

I got an answer of 3.1 m/s thus:
Assuming constant deceleration
\[\frac{v _{2}-v _{1}}{t _{2}-t _{1}}=\frac{v-v _{0}}{t-0}\]
\[-5.6=\frac{v-v _{0}}{4.2}...............(1)\]
Average speed = 62.4/4.2 = speed after 2.1 seconds (4.2/2 seconds)
Speed at t(0) = 14.857 + at = 14.857 + (5.6 * 2.1) = 26.617 m/s
\[v _{0}=26.617 m/s\]
Substituting the value for v(0) into equation (1) gives
\[-(5.6\times 4.2)=v-26.617\]
v = 3.1 m/s

Answer 3

here is given , a = -5.60 m/s^2 , t = 4.2 seconds and s = 62.4 m

since s = ut +\(\frac{1}{2}\)at^2

hence we will put the values of s , a and t here :

\[62.4 = u(4.2)+\frac{1}{2}(-5.60)(4.2)^2\]
\[62.4=4.2 u +49.392\]
\[62.4-49.392-4.2u\]
\[13.008=4.2u\]
\[u = \frac{13.008}{4/2}\]
\[u = 3 (approx.)\]

Answer 4

ami late?

Answer 5

@lgbasallote have u got that solution

Answer 6

why + 49.392?

shouldnt it be negative??

Answer 7

@mathslover

Answer 8

wait for 1 minute @lgbasallote i think i did mistake

Answer 9

wait this is the correct solution now

\[s=ut+\frac{1}{2}at^2\]
\[62.4 = 4.2u -49.392\]
\[62.4+49.392=4.2u\]
\[111.792=4.2u\]
\[26.6=u\]

hence initial velocity = 26.6 m/s

now final velocity : u + at

therefore \(\large{v_f=v_i+at}\) that means \(\large{v_f=26.6 +(-5.6*4.2)}\)

hence \(\huge{v_f=26.6-23.52}\) \(\huge{v_f=3.08 = 3.1ms^{-1}}\)

Answer 10

@lgbasallote did u get it now ?

Answer 11

i mistakenly wrote 6.24 in my solution :O no wonder...

Answer 12

:) np . well sorry for that mistake