Question:
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?
Answer given is 10 seconds.
But equation for distance travelled 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.
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?
Answer given is 10 seconds.
But equation for distance travelled 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.
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Acceleration DOES NOT "tick like a clock". I repeat: Acceleration DOES NOT "tick like a clock".
In otherwords, you cannot go second-by-second analyzing the motion that way.
You need to keep continuity in your approach.
In this particular problem, you are equating a parabola to a line. This will render you solving a quadratic equation.
If you are not familiar solving quadratic equations, I suggest you do some back-reading on the subject.
In otherwords, you cannot go second-by-second analyzing the motion that way.
You need to keep continuity in your approach.
In this particular problem, you are equating a parabola to a line. This will render you solving a quadratic equation.
If you are not familiar solving quadratic equations, I suggest you do some back-reading on the subject.
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x_truck = 1/2 a t^2, a = 2 m/s^2
x_car = v(t-5), v = 20 m/s
1/2 a t^2 = v(t-5) => t^2 = 20(t-5) => t=10 s
You didn't take into account that the car only reaches the point at t=5 s.
x_car = v(t-5), v = 20 m/s
1/2 a t^2 = v(t-5) => t^2 = 20(t-5) => t=10 s
You didn't take into account that the car only reaches the point at t=5 s.