It all depends upon how the object is accelerated.
Assuming a constant acceleration so that it reaches 60 mph after 7 seconds starting from stationary:
1 mile = 5280 ft
1 hour = 3600 s
→ 1 mph = 1 mile / 1 hour = 5280 ft / 3600 s = 22/15 ft/s
v = final_velocity = 60 mph = 22/15 ft/s
t = time = 7 s
u = initial velocity = 0
v = u + at
→ v = 0 + at
→ a = v/t = (22/15) / 7
v² = u² + 2as
→ v² = 0 + 2as
→ s = v² / (2a) ft
→ s = (22/15)² / (2 × (22/15) / 7) ft
→ s = (22/15) / (2/7) ft
→ s = 22/15 × 7/2 = 77/15 ft = 5 2/15 ft ≈ 5.13 ft
If you are asking about a car (or other motorised vehicle), then the acceleration is not going to be constant: the vehicle will start accelerating slowly and then accelerate faster. Depending upon the exact shape of the function of the acceleration with relation to time, will change the distance covered.
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Not enough information. You also need to know at what rate the object in question accelerates.