The driver accelerates from a stop sign, cruises for 20 s at a constant speed of 60 km/h, and then brakes to come to a stop 40 s after leaving the stop sign.
what is the instantaneous acceleration at t=35s?
express your answer using two significant figures.

Description of motion.

1) From the stop sign until a unknowm moment: acceleration from 0 m/s to 60 m/s, during x seconds

You do not know the time during which the driver kept this motion.

2) During 20 s (after having finished of accelerating), constant speed = 60 km/h, from x s until x + 20s

3) From x+20 s to 40 s, c onstant acceleration from 60 km/h to 0 km/h

With the information you cannot tell, in what stage was the driver at t = 35 s, because you need to know x.

Sure that information is in a graph or other part of the question which was not included in your question.

I will tell you how to calculate the acceleration in any of the three stages, once you have x.

1)  First stage:

If the motion is uniformly accelerated, Vf = Vo + at => a = [Vf - Vo] / t

Vf = 60 km/h (which you must pass to m/s)

Vo = 0 (from rest because she started from a stop sign)

t = x seconds

=> a = [60km/h * (1000m/km)*(1h/3600s)] / x = (16.67 m/s) / x s

2) Second stage: constant speed.

Consequently, acceleration = 0

3) Third and last stage: uniform acceleration during 40 - (x+20) s

Once you know x, you will know the time during which she was braking.

Vf = Vo + a*t => a = [Vf - Vo] / t =

a = [0 - 60km/h] * [1h / 3600s] * [ 1000m / 1km] / [40s - (x+20)s]   

a = - [60*1000/3600 m/s] / [20s - xs] =

The negative sign means that the acceleration is opposed to the velocity (braking).
 

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