Answer
$p = (37.5~N\cdot s)~\hat{i}$
Work Step by Step
To find the impulse on the car, we can calculate the area under the force versus time graph.
From $t = 0$ to $t = 7.0~s$, the area can be divided into four parts, including a triangle (0 to 2 s), a rectangle (2 s to 4 s), a triangle (4 s to 6 s), and a triangle (6 s to 7 s)
We can find each area separately:
$A_1 = \frac{1}{2}(10.0~N)(2.0~s) = 10.0~N\cdot s$
$A_2 = (10.0~N)(2.0~s) = 20.0~N\cdot s$
$A_3 = \frac{1}{2}(10.0~N)(2.0~s) = 10.0~N\cdot s$
$A_4 = \frac{1}{2}(-5.0~N)(1.0~s) = -2.5~N\cdot s$
We can find the impulse from $t = 0$ to $t = 4.0~s$:
$J = 10.0~N\cdot s +20.0~N\cdot s+10.0~N\cdot s-2.5~N\cdot s = 37.5~N\cdot s$
The change in the car's momentum will be equal to the impulse. Since the car started at rest, $p = 37.5~N\cdot s$
We an express the momentum in unit-vector notation:
$p = (37.5~N\cdot s)~\hat{i}$
