Please somebody explain this simple physics problem?

In summary: Learning. I'm glad you find it helpful.In summary, the problem is asking for the time it takes for a car, traveling at a constant speed of 20m/s, to overtake a truck that starts with an acceleration of 2m/second square. After 5 seconds, the truck has traveled 25m and the car has traveled 100m, reaching the starting point of the truck. At the 10th second, the truck has traveled 100m and the car has traveled 200m, reaching the truck. However, at this point, the truck's velocity increases to 22m/s while the car's velocity remains at 20m/s. Therefore, according to the given information,
  • #1
mashac
3
0

Homework Statement



A truck start from a place with an acceleration of 2m/second square.A car passes the same place after 5s with uniform speed of 20m/s. Find the time taken in which car overtakes truck?


Homework Equations





The Attempt at a Solution



Answer given is 10 seconds.

But equation for distance traveled by car is d=vt which means d=20*t and for the truck is d=1/2*a*t^2 which means d=t^2.

By this equation after 5s truck will travel a distance of 25m with a velocity of 10 m/s and car will travel 100m and reach the starting point of the truck. At the 10th second the truck will travel a distance of 100m and car will travel a distance of 200m and reach the truck. At this point the velocity of both truck and car will be 20m/s, but car have no acceleration and the truck accelerates at a rate of 2m/second square. So at the 11th second the truck's velocity will be 22 m/s and car's still will be 20 m/s. So according to my stupid theory the car never over take the truck.

I know I'm somehow wrong about this, but please someone point out that exactly which part of my logic is faulty and kindly explain the logic of right answer which is 10 seconds.
 
Physics news on Phys.org
  • #2
But equation for distance traveled by car is d=vt which means d=20*t and for the truck is d=1/2*a*t^2 which means d=t^2.

This isn't right because the car passes the same spot 5 seconds after the truck. So:
d=1/2*a*t^2 for the truck
d=20*(t-5) for the car
 
  • #3
never mind [=> ignore this post]
 
  • #4
Thanks ideasrule, now I understand a little more about the problem.
 
  • #5
An old post for new member !
How Sweet
 

FAQ: Please somebody explain this simple physics problem?

1. What is the problem asking for?

The problem likely involves a physical scenario and is asking for a specific value or relationship between variables.

2. What are the given information and variables?

The problem will provide specific values for certain variables involved in the scenario, and may also list any relevant equations or constants.

3. How do I approach solving the problem?

The problem may require you to use a specific equation or concept from physics. Make sure you understand the given information and the relationships between variables. Then, use appropriate equations and algebraic manipulation to solve for the unknown variable.

4. How do I check if my answer is correct?

Once you have solved for the unknown variable, you can check your answer by plugging it back into the original equation and seeing if it satisfies the given information. Additionally, you can compare your answer to reasonable estimates or known values for similar scenarios.

5. What should I do if I am still stuck?

If you are having trouble understanding the problem or solving it, try breaking it down into smaller parts and tackling them one at a time. You can also consult online resources, textbooks, or a teacher for guidance and clarification.

Similar threads

Replies
9
Views
4K
Replies
2
Views
1K
Replies
36
Views
5K
Replies
4
Views
1K
Replies
25
Views
3K
Replies
2
Views
3K
Replies
10
Views
4K
Back
Top