A robot probe drops a camera off the rim of a 278 m high cliff on Mars, where the free-fall acceleration is 3.7 m/s2 . Find the
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Answer:
The position of the car at t = 1.5 s is at -8.1625 meters
Explanation:
The initial position of the car is 3.2 meters
The initial velocity is -8.4 m/s
The constant acceleration is 1.1 m/s²
We need to find the final position of the car at the time t = 1.5 seconds
The displacement <em>s</em> = final position - initial position
, where <em>u</em> is the initial velocity, <em>a</em> is the
constant acceleration and <em>t</em> is the time
So we can find the final velocity by using the rule:
final position - initial position =
initial position = 3.2 meters , u = -8.4 m/s , a = 1.1 ²m/s , t = 1.5 s
Substitute these values in the rule
final position - 3.2 =
final position - 3.2 = -12.6 + 1.2375
final position - 3.2 = -11.3625
add 3.2 for both sides
final position = -8.1625
<em>That means the car is at 8.1625 meters in opposite direction</em>
<em>The position of the car at t = 1.5 s is at -8.1625 meters </em>
Answer:
Final speed of the car, v = 24.49 m/s
Explanation:
It is given that,
Initial velocity of the car, u = 0
Acceleration,
Time taken, t = 7.9 s
We need to find the final velocity of the car. Let it is given by v. It can be calculated using first equation of motion as :
v = 24.49 m/s
So, the final speed of the car is 24.49 m/s. Hence, this is the required solution.