A car traveling at 95 km/h strikes a tree. The front end of the car compresses and the driver comes to rest after traveling 0.80 m. What was the magnitude of the average acceleration of the driver during the collision?
Answers
Answered by
0
Answer:
The acceleration is
=
−
348.4
m
s
−
2
Explanation:
The initial velocity of the car (driver) is
u
=
85
k
m
h
−
1
u
=
85
3.6
=
23.61
m
s
−
1
The distance travelled by the driver is
s
=
0.80
m
The final velocity of the driver is
v
=
0
m
s
−
1
Apply the equation of motion,
v
2
=
u
2
+
2
a
s
The acceleration is
a
=
v
2
−
u
2
2
s
=
0
−
23.61
2
2
⋅
0.80
=
−
348.4
m
s
−
2
Answered by
1
hey mate here is your answer
Convert velocity to meters per second:
95km/h * 1m/s / 3.6km/h = 26.4 m/s
v² = u² + 2as
Final velocity is 0, so
a = u² / 2s = (26.4m/s)² / 2*0.90m = 387 m/s² ◄
387 / 9.8 = 39 "g"s ◄
hope it helps you dear..... ⭐⭐⭐⭐
Convert velocity to meters per second:
95km/h * 1m/s / 3.6km/h = 26.4 m/s
v² = u² + 2as
Final velocity is 0, so
a = u² / 2s = (26.4m/s)² / 2*0.90m = 387 m/s² ◄
387 / 9.8 = 39 "g"s ◄
hope it helps you dear..... ⭐⭐⭐⭐
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