solve this ....
now i am getting tired
what i will do
u tell
Answers
Answer:
Let d be the distance between the two cars at t=0t=0, which is 220m220m, and the velocity, v_1v1 be 20km/h20km/h which in m/sm/s is 5.55m/s5.55m/s. Let v_2v2 be the 40km/h40km/h which is 11.1m/s11.1m/s.
To solve this problem, we will simultaneously solve two equations at once to find the initial velocity and the acceleration.
The two equations we will use are:
d-x_1=v_0t_1+\frac{1}{2}at_1^2d−x1=v0t1+21at12 where t_1=\frac{x_1}{v_1}t1=v1x1
and
d-x_2=v_0t_2+\frac{1}{2}at_2^2d−x2=v0t2+21at22 where t_2=\frac{x_2}{v_2}t2=v2x2
Subsisting x_1=44.5mx1=44.5m and x_2=76.7mx2=76.7m and our other given values leads us to the following results:
v_0=-13.9 m/sv0=−13.9m/s and in km/h we have -50km/h.
and the acceleration is, a=-2.0m/s^2a=−2.0m/s2.
As both values are negative, it tells us that it is along the -x direction
Step-by-step explanation:
Hope this helps you!!
Please mark as brainlist!!
Please
Let dd be the distance between the two cars at t=0t=0, which is 220m220m, and the velocity, v_1v
1
be 20km/h20km/h which in m/sm/s is 5.55m/s5.55m/s. Let v_2v
2
be the 40km/h40km/h which is 11.1m/s11.1m/s.
To solve this problem, we will simultaneously solve two equations at once to find the initial velocity and the acceleration.
The two equations we will use are:
d-x_1=v_0t_1+\frac{1}{2}at_1^2d−x
1
=v
0
t
1
+
2
1
at
1
2
where t_1=\frac{x_1}{v_1}t
1
=
v
1
x
1
and
d-x_2=v_0t_2+\frac{1}{2}at_2^2d−x
2
=v
0
t
2
+
2
1
at
2
2
where t_2=\frac{x_2}{v_2}t
2
=
v
2
x
2
Subsisting x_1=44.5mx
1
=44.5m and x_2=76.7mx
2
=76.7m and our other given values leads us to the following results:
v_0=-13.9 m/sv
0
=−13.9m/s and in km/h we have -50km/h.
and the acceleration is, a=-2.0m/s^2a=−2.0m/s
2
.
As both values are negative, it tells us that it is along the -x direction.