Question

Alexander's hobby is dirt biking. On one occasion last weekend, he accelerated from rest to 17.8 m/s in 1.56 seconds. He then maintained this speed for 9.47 seconds. Seeing a coyote cross the trail ahead of him, he abruptly stops in 2.79 seconds. Determine Alexander's average speed for this motion.

@Vaidehi09

Answer 1

So the issue i'm having, is the last part, where u said vf=vo-at...am i finding vf in this or acceleration?

Answer 2

i will repost the answer u posted before.

Answer 3

we know that avg speed = (total distance traveled) / (total time take).........(i)
now total time = ( 1.56 + 9.47 + 2.79 ) secs.
the total distance covered is calculated as follows:
the whole journey can be divided into three parts:
1) when he accelerated from rest
2) when he moves with constant speed
3) when he decelerated and came to a stop.
now, find the distance covered in each of these parts using the kinematic equations.

1) given: u1 =0, a1 = 17.8m/s^2, t1 = 1.56 sec
s1 = u1t1 + 1/2 a1 t1^2
s1 = 0 + 1/2 (17.8) (1.56)^2
s1 = 8.9 (1.56)^2 m

also, the constant speed acquired v1 = u1 + a1 t1
v1 = (17.8) (1.56) m/s

2) given: constant speed v1 = (17.8)(1.56) m/s, t2 = 9.47 sec
s2 = v1 * t2
s2 = (17.8) (1.56) (9.47) m

3) given: v3 = 0, t3 = 2.79 s, v1 = u3 = (17.8)(1.56) m/s
first find the deceleration [-a3]
v3 = u3 - a3 t3
u3 = a3 t3
a3 = u3/ t3
a3 = (17.8) (1.56) / (2.79) m/s^2

(v3)^2 = (u3)^2 - 2 a3 s3
(u3)^2 = 2 a3 s3
s3 = (u3)^2 / 2 a3
s3 = (17.8)^2 (1.56)^2 (2.79) / 2 (17.8) (1.56)
s3 = (17.8) (1.56) (2.79) / 2
s3 = 8.9 (1.56) (2.79) m

so, total distance covered = s1 + s2 + s3.
substitute this in (i) and u'll get the avg speed.

The Changes:


1) given: u1 = 0, v1 = 17.8, t1 = 1.56
use v1 = u1 + a1 t1.......to find a1.
then sub that s1 = 1/2 a1 t1^2......to get s1.

2) given: constant speed = v1 = 17.8, t2 = 9.47
calculate s2 using, s2 = v1 t2

3) here, proceed the original way. the only change will be,
v1 = u3 = 17.8

Answer 4

u already know that Vf = 0. and Vo is given. so first find out a.

Answer 5

6.38m/s^2...so this would be substituted into d=vot+1/2at^2?

Answer 6

yeah, u can do that. but if u use v^2 = u^2 + 2as, it'll be easier since v = 0.

Answer 7

ok so i did the calculations and i added up all the distances. 1193.18meters, which i have to divide by 13.82, which is the total time. are the numbers correct?

Answer 8

oh, i didn't calculate all that. give me a sec.

Answer 9

i got the total distance = 207.28m

Answer 10

I got that before, but the answer on the website is 23.4m/s...if you divide 207.28 by 13.82, you get around 15m/s.

http://www.physicsclassroom.com/calcpad/1dkin/problems.cfm

number 32.

Answer 11

u heard the guided solution? its not opening on my comp.

Answer 12

the plugin doesn't exist anymore...so yea.

Answer 13

btw, how did u get 1193.18 ?

Answer 14

umm. i got 1010.72m for the last one with:

d=Vo^2/2a
d=17.8^2/2(6.38)
d=1010.72m.

Then i added the rest.

Answer 15

oh never mind, my calculator did it wrong.

Answer 16

24.83m

Answer 17

Yea so using that i get 207.29m

Answer 18

ah! got it. to get 24 as answer....proceed the the way i showed in the beginning. don't see the correction part. coz in the question on that site, 17.8 is given as acceleration and not velocity.

Answer 19

ohhhhhhh i seeee. ok so i was correct the first time around...lol i'm so dumb. THANKS A TON! :)

Answer 20

lol! sure, not a prob! :)