Question

A car and a motorcycle start from rest at the same time on a straight track but the motorcycle is 25.0 m behind the car, Figure 1. The car accelerates at a uniform rate of 3.70 m/s2 and the motorcycle at a uniform rate of 4.40 m/s2. (a) How much time elapses before the motorcycle overtakes the car? (b) How far will each have traveled during that time?

Answer 1

ok , do u know relative velocity ?

Answer 2

@Deeza

Answer 3

yes

Answer 4

so what is the relative velocity of the motorcycle with respect to the car ?

Answer 5

and the car distance 227.10 m
the motorcycle would have a distance of 240.34
The difference between the two distances is 13.24 m
time elapses 8.452 seconds

Answer 6

(a) How much time elapses before the motorcycle overtakes the car?

This will be when they've travelled equal distance.

Use some physics motion equations (d = vt + at^2), they both start from rest so v=0, so:

\[\Large \text{car distance} = 3.7t^2 + 25\] (add 25since it starts 25m ahead)

\[\Large \text{bike distance }= 4.4t^2\]
Set them equal to each other and solve for t.

Answer 7

did you get the solution?

Answer 8

yeah, thank you all

Answer 9

What was your solution (and do you understand it)?

Answer 10

my solution was
a) motocycle
X = X0 + VO T +1/2 AT^2
X = 0 + 0t + 1/2 (4.40t^2)
x = 2.2t^2

Car
x = 25 + 1/2 (3.70)t^2
x= 25 + 1.85t

motorcycle = car
2.2t^2 = 25 + 1.85t^2
2.2t^2 - 25 - 1.85t^2 = 0
0.35t^2 - 25 = 0
0.35t^2/0.35 = 25/0.35
t^2 = 25/0.35
t= 8.45 seconds

b) motorcycle
x +x0 + v0t + 1/2 at^2
x= 0+0(8.45) + 1/2(4.40)(8.45)^2
=157m

Car
x = x0 + v0t + 1/2 at^2
= 25 +0(8.45) + 1/2 (3.70)(8.45)^2
=157m

Answer 11

Good job, i just realized i forgot the 1/2 in the d = vt + at^2, oops

Answer 12

lol! yeah

Answer 13

What did you need our help for then :P seems like you had it

Answer 14

i was lost at first but your ideas helped me:)

Answer 15

oh good :)