A )
t 1 = 2 h, t 2 = 6 h
Δ t = t 2 - t 1 = 6 - 2 = 4 h
54 = 50 + a Δ t
54 = 50 + 4 a
4 a = 54 - 4
4 a = 4
a = 4 : 4
a = 1 km/h²
v o = 48 km/h
An equation that can be used to describe the velocity of the car at the different times is:
v = 48 + t
B ) The graph is in the attachment.
Answer:
22.5 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 30 m/s
Time (t) = 1.5 s
Final velocity (v) = 0 m/s
Distance (s) =?
The distance to which the car move before stopping from the time the driver applied the brake can be obtained as follow:
s = (u + v)t/2
s = (30 + 0)1.5 / 2
s = (30 × 1.5) / 2
s = 45 / 2
s = 22.5 m
Thus, the car will move to a distance of 22.5 m before stopping from the time the driver applied the brake.
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