Question Video: Finding the Momentum of an Object at a Given Time given Its Displacement with Respect to Time | Nagwa Question Video: Finding the Momentum of an Object at a Given Time given Its Displacement with Respect to Time | Nagwa

Question Video: Finding the Momentum of an Object at a Given Time given Its Displacement with Respect to Time Mathematics • Third Year of Secondary School

A car of mass 1,350 kg moves in a straight line such that at time 𝑡 seconds, its displacement from a fixed point on the line is given by 𝑠 = (6𝑡² − 3𝑡 + 4) m. Find the magnitude of the car’s momentum at 𝑡 = 3𝑠.

03:13

Video Transcript

A car of mass 1350 kilograms moves in a straight line such that, at time 𝑡 seconds, its displacement from a fixed point on the line is given by 𝑠 equals six 𝑡 squared minus three 𝑡 plus four meters. Find the magnitude of the car’s momentum at 𝑡 equals three seconds.

We’re looking for the magnitude of the car’s momentum. Momentum is a course of vector quantity. But we’re only interested in the magnitude of this quantity. Now you might know that the momentum of an object is the product of its mass and velocity. And this is good news because we know the mass of the car. It’s 1350 kilograms.

But we’re not explicitly told the velocity of the car anywhere in the question. What we are told is its displacement 𝑠 equals six 𝑡 squared minus three 𝑡 plus four meters, where 𝑡 is the time in seconds. And we can use this displacement to find the velocity as the velocity at any given point in time is just the instantaneous rate of change d by d𝑡 of the displacement at that point in time.

So we can call the velocity 𝑣 and displacement 𝑠 as it is in the question. And we can substitute the expression we have for 𝑠 in terms of 𝑡. We’re told that the displacement 𝑠 is six 𝑡 squared minus three 𝑡 plus four. And we can differentiate term by term.

The derivative of 𝑡 squared with respect to 𝑡 is two 𝑡. And so the derivative of six 𝑡 squared is six times two 𝑡, which is 12𝑡. The derivative of 𝑡 with respect to 𝑡 is just one. And so the derivative of three 𝑡 with respect to 𝑡 is three. And the derivative of a constant is zero. And so the constant term four doesn’t contribute anything to the velocity.

We found therefore that the velocity at time 𝑡 is 12𝑡 minus three. And as the displacement was measured in metres and the time in seconds, this velocity has units of metres per second.

So now that we have the velocity at time 𝑡, we can substitute it in to our equation for momentum. The momentum is 1350 times 12𝑡 minus three. And as the mass was measured in kilograms and the velocity was measured in metres per second, this momentum is measured in kilograms metres per second. This is the momentum at anytime 𝑡. But we’re only interested when 𝑡 is three seconds.

So we substitute 𝑡 equals three here. 𝑡 is three and so 12𝑡 is 36. And probably, the most sensible thing to do is just to put this into our calculator to get a momentum of 44550 kilogram metres per second. That’s the momentum of our car after three seconds. And as this car is moving in a straight line and the momentum turned out to be positive, it’s also the magnitude of the car’s momentum.

This is therefore the answer to our question, 44550 kilogram metres per second.

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