Answer:
150.5 m
Explanation:
The motion of the car is a uniformly accelerated motion, so we can use the following suvat equation to find the displacement:
where
s is the displacement
u is the initial velocity
t is the time
a is the acceleration
For the car in this problem,
u = 0
Substituting t = 8.9 s, we find the displacement of the car at that time:
Now
For the second question
Using 2nd equation of motion
a) Vector product: |d1 × d2| = (34.2 m) k
b) Scalar product: d1 · d2 = -0.874 m
c) (d1 + d2) · d2 = 16.1 m
d) Component of d1 along direction of d2: -0.21 m
Explanation:
a)
In this part, we want to calculate
|d1 × d2|
Which is the vector product between the two displacements d1 and d2.
The two vectors are:
d1 = (6.00 m)i + (5.74 m)j
d2 = (-2.92 m)i + (2.9 m)j
The vector product of two vectors and
is also a vector which has components:
We notice immediatly that in this problem ,the two vectors d1 and d2 lie in the x-y plane, so they do not have components in zero. Therefore, the vector product has only one component, which is the one in z, and it is:
Therefore, the vector product of d1 and d2 is:
|d1 × d2| = (34.2 m) k
b)
In this case, we want to calculate
d1 · d2
Which is the scalar product between the two displacements.
The scalar product of two vectors and
is a scalar given by:
In this problem,
d1 = (6.00 m)i + (5.74 m)j
d2 = (-2.92 m)i + (2.9 m)j
Therefore, the scalar product between the two vectors is:
c)
In this case, we want to calculate
(d1 + d2) · d2
Which means that first we have to calculate the resultant displacement d1 + d2, and then calculate the scalar product of the resultant vector with d2.
Given two vectors and
, the resultant vector is also a vector given by
In this case,
d1 = (6.00 m)i + (5.74 m)j
d2 = (-2.92 m)i + (2.9 m)j
So the resultant vector is
So
And calculating the scalar product with d2, we find:
d)
The component of a vector a along another vector b is given by
where is the scalar product between and b
is the magnitude of vector b
In this problem, we have the two vectors
d1 = (6.00 m)i + (5.74 m)j
d2 = (-2.92 m)i + (2.9 m)j
We want to find the component of d1 along the direction of d2.
We already calculated the scalar product of the two vectors in part b):
d1 · d2 = -0.874 m
The magnitude of a vector b is given by
So, for vector d2,
Now we can calculate the component of d1 along d2:
Learn more about operations with vectors:
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