Question

At a stoplight, a truck traveling at 15 m/s passes a car as it starts from rest.?

Answer 1

i dont know how to combine the equations

Answer 2

what u wanna find ???

Answer 3

At a stoplight, a truck traveling at 15 m/s passes a car as it starts from rest. The truck travels at constant velocity and the car accelerates at 3 m/s^2. How much time does the car take to catch up to the truck?

Answer 4

I want to know how to solve it step by step :\

Answer 5

i know that some how we get to this but i dont understand how, 15t=1/23t^2

Answer 6

15t=(1/2)3t^2

Answer 7

so find t

Answer 8

ohh srry

Answer 9

i dont know how to get that though?
i copied it from the board the teacher didnt explain it

Answer 10

u see car moves with constant velocity . u let t be d time elapsed in which d whole thing takes place n as we can see that the displacement of car {(1/2)3t^2} n truck(15t) will be equal

Answer 11

therefore (1/2)3t^2= 15t

Answer 12

is there a formula that you used to substitute the initial velocity and acceleration

Answer 13

see when there is constant acceleration then we have 3 eqns
v = u +at
s=ut+1/2at^2
2as=v^2-u^2
where u is initial velocity , v is final velocity s is displacement n a is cnstant acceleration

Answer 14

i used d 2nd one 4 finding displacement of car in time t

Answer 15

hello is it clear ?????

Answer 16

hey sorry

Answer 17

hey dude r u satisfied with d ans n sorry stands 4 what??????

Answer 18

oh i was away yeah ill figure it out, my teacher uses different symbols fir final velocity and initial and displacement

Answer 19

kk fine u just put d values n symbols do not matter anyways ,,,,,,,, bbyeeeee

Answer 20

Well explain

Answer 21

The car will overtake the truck when it catches up. The technical way of saying that is that they will both have gone the same distance from the traffic lights. That's it. that's what overtaking means.

The distance one goes under constant acceleration is described by the equation given above by arzoo: s (distance) = u(initial velocity)*t(time) + 1.2 * a(acceleration) * t(time)^2

The truck doesn't accelerate but just goes at a constant speed. So its a is zero and its distance s is
s(truck) = u*t = 15t

The car starts at u=zero but does accelerate. So its distance is
s(car)= 1/2 * a * t^2 = 1/2 * 3 * t^2

And these are equal when the car overtakes - because it has caught up and the distances are the same. So...

s(truck)=s(car)
15t = 1/2 * 3 * t^2

Rearrange and cancel out... Just as arzoo says.