Video Transcript
A car that was initially moving at
a steady speed travels a distance of 15 meters while accelerating in a straight line
for 10 seconds at 2.5 meters per second squared in the opposite direction to its
original velocity. What was the car’s initial
speed?
Okay, so in this question, we’ve
got a car that we’re told is initially moving at a steady speed. Right now, we don’t know what that
speed is. And in fact, that’s exactly what
the question is asking us to work out. In this sketch, we’ve drawn the car
moving to the right. So we can say that the car has an
initial velocity that’s directed to the right, and we’ll label the magnitude of this
initial velocity as 𝑢.
Recall that velocity is a vector
quantity, and that means that as well as a magnitude, it has a direction. In this case, the magnitude of the
velocity is 𝑢, and its direction is to the right. Now, the question is asking us for
the car’s initial speed, and we can think of speed as being like the scalar
counterpart to velocity. Since speed is a scalar quantity,
it doesn’t have a direction associated with it and instead is completely defined by
a magnitude. In the case of the car from this
question, we have defined 𝑢 as the magnitude of the initial velocity, and that will
also be equal to the car’s initial speed.
We might wonder why it’s even worth
talking about the vector quantity velocity when we’re only asked to find the car’s
initial speed. The reason is that directional
information is going to be important in this question, and in fact we’re told that
the car accelerates in the opposite direction to its original velocity. So since the velocity is directed
to the right, that means that the acceleration, which we’ve labeled as 𝑎, must be
directed to the left.
Whenever the acceleration of an
object is in the opposite direction to the object’s motion, that acceleration acts
to slow the object down, and in this case we could also call it a deceleration. We’re told that the acceleration
has a magnitude of 2.5 meters per second squared. And so we’re going to say that this
quantity 𝑎, the acceleration of the car, has a value of negative 2.5 meters per
second squared. Now this negative sign comes in
because acceleration is a vector quantity, just like velocity. So it also has a direction. By talking about the quantity 𝑢 as
being the magnitude of the initial velocity, we were implicitly taking this
rightward direction of the initial velocity to be the positive direction.
If right is the positive direction,
then this means that all vector quantities directed to the right will have a
positive value, while all vectors directed to the left will be negative. Since the acceleration of the car
is directed to the left, then that means that the acceleration’s value will have a
negative sign. We’re told that the car accelerates
for a time of 10 seconds. And we’ve labeled this time as
𝑡. We’re also told that the car
travels a distance of 15 meters, while it accelerates. So after this 10-second interval
during which the car accelerates has passed, it will have moved a distance of 15
meters, which we’ve labeled as 𝑠.
So we know the value of the car’s
acceleration, we know how much time it accelerates for, and we know the distance
that it travels while it accelerates. What we don’t know is the car’s
initial velocity. Luckily for us though, it turns out
that one of the kinematic equations links these four quantities together. Specifically, the equation that we
want says that 𝑠 is equal to 𝑢 times 𝑡 plus a half times 𝑎 times 𝑡 squared.
There are two conditions that have
to apply in order to be able to use this equation. The first is that the motion has to
be in a straight line, and the second is that the acceleration must have a constant
value. In our case, we’re told that the
acceleration occurs in a straight line, so we know that the first condition is
met.
We’ve also got a value for this
acceleration of negative 2.5 meters per second squared. The magnitude and the direction of
this are fixed quantities because neither of them depend on the time or the distance
traveled. This means that the acceleration is
indeed constant.
Since both conditions are met, this
means that we can safely go ahead and use this equation to help us solve this
problem. Now, in our case, the quantity that
we’re trying to find out is this initial velocity 𝑢, so we want to rearrange this
equation in order to make 𝑢 the subject. To do this, we will begin by
subtracting a half times 𝑎 times 𝑡 squared from both sides of the equation. On the right, these two terms, so
that’s plus a half times 𝑎 times 𝑡 squared and minus a half times 𝑎 times 𝑡
squared, cancel each other out.
From here, we can then divide both
sides of the equation by the time 𝑡. And then on the right-hand side the
𝑡 in the numerator cancels with the 𝑡 in the denominator, leaving 𝑢 all by
itself. We can then write this equation the
other way around to say that 𝑢 is equal to 𝑠 minus a half times 𝑎 times 𝑡
squared all divided by 𝑡. If we write the right-hand side as
two separate fractions, then we can see that in this second fraction one factor of
𝑡 can cancel from the numerator and denominator. We can then rewrite this right-hand
side as 𝑠 over 𝑡 minus 𝑎 times 𝑡 over two.
Now that we’ve got an equation
where the initial velocity 𝑢 is the subject, we can take our values for 𝑠, 𝑎, and
𝑡 and substitute them in. When we do that, we end up with
this expression here for 𝑢. The first term is 15 meters, that’s
our value for 𝑠, divided by 10 seconds, our value for 𝑡. The first term works out as 1.5
meters per second. Then we subtract from this a second
term equal to negative 2.5 meters per second squared, that’s the acceleration 𝑎,
multiplied by 10 seconds, the time 𝑡, all divided by two. This second term comes out as
negative 12.5 meters per second. Overall then, we have that 𝑢 is
equal to 1.5 meters per second minus negative 12.5 meters per second.
Since we’re subtracting a negative
quantity, these two minus signs negate each other. So then we have that 𝑢 is equal to
1.5 meters per second plus 12.5 meters per second. And this works out as 14 meters per
second. In terms of an initial velocity,
then, as we’ve drawn things in our sketch, we could say that the car’s initial
velocity is either positive 14 meters per second or 14 meters per second to the
right. However, we’re being asked about
the car’s initial speed, which is the magnitude of its initial velocity. That means we don’t want to include
any directional information with our answer. Our answer to this question is
therefore that the initial speed of the car was 14 meters per second.
