physics homework problem help plzzz?

Question

driving at 30 m/s susan enters a one-lane tunnel, and
155 m ahead of her, a van is going at 5 m/s
she deaccelerates at -2 m/s
1. will there be a collision?
2. if so determine how far into the tunnel it will be and at what time it occurs
3. if not determine the distance between susan and the van

please show all work and use one of the five kinematics acceleration equations..thanks!!!

Answer 1

ds=S2-S1 how close they will come for the same period of time t

S1= V1t (Van's equation)
S2= V0t - 0.5 a t^2 (Susan's equation)
V2= V0 - at (this is the speed Susan will have at the time t)
V2<=V1 (just for a miss or 'kiss' )
t>=(V0-V2 )/a=(30-5)/2=12.5 s
S1=V1 t=5 x 12.5=62.5m
S2=V0t - 0.5 a t^2
S2=30 x 12.5 - 0.5 x 2 x (12.5)^2
S2=218.75 m

dS=S2- S1= 218.7 -62.5m= 156.25 m

since dS>155 m she will tap the van.

1. it seems that she must brake a bit harder
2. 218.7 m into the tunnel
3. ....

Answer 2

the kinetic capability = a million/2 mv^2 so on your merchandise you have 2 gadgets that are reported flying aside with equivalent velocity. that's risk-free to anticipate from thermodynamics that an equivalent and opposite reaction has befell with the 0.33 piece. you may desire to interrupt those down into the vertical and horizontal aspects of the action. (you will locate for the two reported products their horizontal action is equivalent and opposite) which leaves in common terms the vertical factor for the final piece. you will locate that the horizontal action in meters in line with 2d it relatively is 15*sin(25) and -15sin(25) respectively (for joules of capability those motions the two count form and you may desire to take genuinely the fee of the two a sort of while calculating.) so as quickly as I calculate the quantity of capability required to advance up each and each of those products i'm getting a million/2 * 3 * (15*sin(25))= 60.2796 joules so for the horizontal action (of the two products) you desire one hundred twenty.5592 joules to push them for the vertical factor (additionally in meters in line with 2d) is 15*cos(25) for each piece. it relatively is 13.5946 m/s so a million/2 * mv^2 = 277.2204 joules (for each piece) so we've 554.4408 joules of capability pushing each and each of those products vertically. via fact we anticipate explosions take place in vertical and horizontal guidelines evenly (yet no longer unavoidably the comparable in horizontal via fact the vertical course) (relatively we anticipate that it pushed west as perplexing because it pushed east even though it ought to have pushed harder interior the southern and northerly guidelines) all of us understand the final 3 Kg piece became into pushed with 554.4408 joules of capability. so we could desire to take the summation of forces (sigma E sub F) the summation of the forces = one hundred twenty.5592 +277.2204 + 277.2204 + 554.4408 = 1229.4408 joules of capability finished. desire to truly galvanize your professor? all of us understand that the third piece weighed 3kg, and all of us understand that the piece became into pushed with 554.4408 joules of capability and it became into on a trajectory one hundred fifty five ranges remote from each and all of the different products. 554.4408 joules = a million/2 * 3* v^2 we remedy that by utilising shifting the 0.5 and the three over to the different area and taking the sq. root. the final piece became into shifting at 19.2257 m/s^2 wish this helped you.