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Drupady [299]
3 years ago
6

A car is going 3 m/s, then it speeds up for 4 seconds. If it has an acceleration of 5m/s^2, find its final velocity. Which equat

ion can you use?
A Δx = vi*t + (½)at^2
B vf = vi + a*t
C Δx = (½)(vf + vi)*t
D vf^2 = vi^2 + 2aΔX
Physics
1 answer:
andreyandreev [35.5K]3 years ago
4 0

Well, since the question GIVES you the initial velocity, the acceleration, and the time, and ASKS for the final velocity, you'd be smart to find an equation that USES the initial velocity, the acceleration, and the time, and FINDS the final velocity.

Have a look at equation B ..... vf = vi + a*t . That's pretty durn close !

vf = (initial velocity) + (acceleration)*(time)

vf = (3 m/s) + (5 m/s²)*(4 sec)

vf = (3 m/s) + (20 m/s)

vf = 23 m/s

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To stop a car, first you require a certain reaction time to begin braking; then the car slows under the constant braking deceler
ANEK [815]

Answer:

a) t_r = 0.55 s

b) a = 5.59 m/s²

Explanation:

given,

total distance traveled by the car to stop is 56.9 m when speed of vehicle is 80 km/h or 80 × 0.278 = 22.24 m/s

total distance traveled by the car to stop is 25.7 m when speed of vehicle is 50.7 km/h or 50.7 × 0.278 = 14.09 m/s

using stopping distance formula

s_1 = v_1 t_r +\dfrac{v_1^2}{2 a}................(1)

s_2 = v_2 t_r +\dfrac{v_2^2}{2 a}..............(2)

on solving both the equation we get

a = \dfarc{v_1v_2(v_1-v_2)}{2(s_1v_2-s_2v_1)}

a = \dfarc{22.4\times 14.09(22.24-14.09)}{2(56.9\times 14.09-25.7\times 22.24)}

a = 5.59 m/s²

now reaction time calculation

t_r =\dfrac{v_1^2d_2-v_2^2d_1}{v_1v_2(v_1-v_2)}

t_r =\dfrac{22.24^2\times 25.7-14.09^2\times 56.9}{22.4\times 14.09(22.24-14.09)}

t_r = 0.55 s      

8 0
3 years ago
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