Physics, asked by 22onverm, 11 months ago

A professional race-car driver buys a car that can accelerate at 6.4 m/s^2. The racer decides to race against another driver in a souped-up stock car. Both start from the rest, but the stock-car driver leaves 1.2 s before the driver of the sports car. The stock car moves with a constant acceleration of +3.3 m/s^2. Find the time it takes the sports-car driver to overtake the stock-car driver.
The answer is in units of s.

Answers

Answered by geniusalbert
0

Answer:

9m/s² is the answer for the question

Answered by ssonu43568
0

Answer:

Unit value is 1.2S.

Explanation:

The physics eqn is just

x = acc*t2/2 + v*t + x0

Displacement x0

So

x = acc*t2/2

Car a has acc = 6.4

But leaves 1.2 sec

after car b

car b has acc = 3.3

xb = 3.3*t2/2 for car b

xa = 6.4*ta2/2 for car a

(use ta = t-1.2)

xa = 6.4*(t-1.2)2/2

Expanding xa = 6.4* (t2 -3t + 9/4) / 2

now just set

xa = xb

and

Solve for t (time which car b starts)

3.3*t2/ 2 = 6.4*(t2 -3t + 9/4) / 2 mult out the 2 on each side

3.3*t2 = 6.4*t2 - 16.5t + 12.375

Now subt

3.3*t2

From each side

1.1*t2 - 16.5t + 12.375 = 0 a = 1.1

b = -16.5

c = 12.375

Solve by quadratic eqn of form ax2 + bx + c

t = -(-16.5) +/- sqrt( (-16.5)2 - 4*1.1*12.375) ) / (2*1.1)

t = (16.5 +/- 14.75805 ) / 2.2

Now, if we use the minus part, or

(16.5 - 14.758)/2.2

t= 0.792

This is not valid,

Since at

t=1.2s

When car a starts,

So the unit value is sec.

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