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Question: A car moving with constant acceleration covered the distance between two points \(60.0m\) apart in \...

A car moving with constant acceleration covered the distance between two points 60.0m60.0m apart in 6.00s6.00s . Its speed as it passed the second point was 15.0m/s15.0m/s . What was the speed at the first point?

Explanation

Solution

Here, we are given the final speed, distance covered, and the time taken to cover the distance. So, using the two equations, v=u+atv=u+at and v2=u2+2as{{v}^{2}}={{u}^{2}}+2as , we can find the speed at the first point, i.e., the value of uu.

Complete step by step solution:
We are given, vv = speed of the car at the second point =15m/s=15m/s ,
ss = distance between the two points =60m=60m ,
tt = time of motion =6s=6s
Now, let, uu= speed of the car at the first point,
And aa = acceleration.
From the equation of motion, we know, v=u+atv=u+at
So, 15=u+6×a15=u+6\times a
Therefore, a=15u6a=\dfrac{15-u}{6} ……………………(i)
Now, we also know that, v2=u2+2as{{v}^{2}}={{u}^{2}}+2as
Therefore, using the value of aa from the equation (i), we can have,
225=u2+(15u)6×60\Rightarrow 225={{u}^{2}}+\dfrac{(15-u)}{6}\times 60
225=u2+30020u\Rightarrow 225={{u}^{2}}+300-20u
u2+7520u=0\Rightarrow {{u}^{2}}+75-20u=0
u215u5u+75=0\Rightarrow {{u}^{2}}-15u-5u+75=0
Now, solving this equation, we can have, (u15)(u5)=0(u-15)(u-5)=0
Therefore, u=15u=15 and u=5u=5 .
As, the value of vv , i.e., the speed of the car at the second point is 15m/s15m/s ,
So, uu = the speed of the car at the first point =5m/s=5m/s.

Additional information:
If a particle is moving in a straight line, in a three-dimensional space, with constant acceleration, the three equations of motion can be applied to them to calculate the position, or speed, or acceleration of the particle. The three equations of motion are:
v=u+atv=u+at
s=ut+12at2s=ut+\dfrac{1}{2}a{{t}^{2}}
v2=u2+2as{{v}^{2}}={{u}^{2}}+2as
Where
uu is the initial speed of the particle
vv is the final velocity of the particle,
ss is the displacement of the particle,
aa is the acceleration of the particle (in case of the bodies moving under the influence of gravity, the standard gravity gg is used)
And, tt is the time interval.

Note: These three equations of motion are often referred to as the SUVAT equations, where the word “SUVAT” is an acronym from the variables, ss= displacement, uu = initial speed, vv = final speed, aa = acceleration, and tt = time.