In the given figure, ST is a straight line and ray QP stands on it.
∴ \(∠\)PQS + \(∠\)PQR = 180º (Linear Pair)
\(∠\)PQR = 180º − \(∠\)PQS............ (1)
\(∠\)PRT + \(∠\)PRQ = 180º (Linear Pair)
\(∠\)PRQ = 180º − \(∠\)PRT............ (2)
It is given that \(∠\)PQR =\(∠\) PRQ.
Equating equations (1) and (2), we obtain
180º − \(∠\)PQS = 180º− \(∠\)PRT
\(∠\)PQS = \(∠\)PRT
A driver of a car travelling at \(52\) \(km \;h^{–1}\) applies the brakes Shade the area on the graph that represents the distance travelled by the car during the period.
Which part of the graph represents uniform motion of the car?