Question

Physics question:

A car traveling at 35.0 km/h speeds up to 45.0 km/h in a time of 5.00s. The same car later speeds up from 65.0 km/h to 75.0km/h in a time of 5.00s. (a) Calculate the magntude of the constant acceleration for each of these intervals. (b) Determine the distance traveled by the car during each of these time intervals.

Provide the formula and answers for both questions.

Answer 1

Here ,

a) for the first interval ,

u = 35 km/hr

u = 9.72 m/s

v = 45 km/h

v = 12.5 m/s

acceleration = (v - u)/t

acceleration = (12.5 - 9.72)/5

acceleration =0.556 m/s^2

the acceleration of the car is 0.556 m/s^2

Now, for the second interval .

acceleration = (75 - 65) * 1000/(3600 * 5)

acceleration = 0.556 m/s^2

the acceleration for this interval is 0.556 m/s62

b)

distance during first interval

Using second equation of motion

d = u*t + 0.5 * at^2

d = 9.72 * 5 + 0.5 * 0.556 * 5^2

d = 55.55 m

the distance travelled during first interval is 55.6 m

for the second interval ,

d = 65 * 1000 * 5/3600 + 0.5 * 0.556 * 5^2

d = 97.23 m

the distance travelled during this interval is 97.23 m